WORK AND ENERGY
Very Short Answer Type Questions
Q1) How much work.................... m is raised to a height h above the ground ?
Sol. Work done = m × g × h , where g is acceleration due to gravity.
Q2) State.................... of work.
Sol. Joule.
Q3) Is work.................... quantity ?
Sol. Scalar.
Q4) Define 1 Joule ....................
Sol. 1 Joule of work is done when 1 N force moves a body by 1 m.
Q.5 What is the condition.................... on a body ?
Sol. The force should produce motion in the body.
Q6) Is energy.................... quantity ?
Sol. No.
Q7) What a.................... of (a) work, and (b) energy ?
Sol. (a) Joule (b) Joule
Q8) What is the work.................... along a friction less surface?
Sol. Zero.
Q9) By how much.................... if its speed is doubled?
Sol. The kinetic energy of a body will become four times.
Q10) Write an expression ....................mass m moving with a velocity v.
Sol. 1/2 mv2
Q11) If the speed of a body is halved,.................... its kinetic energy ?
Sol. It becomes one-fourth
Q12) On .................... kinetic energy of a body depend ?
Sol. It depends on mass and velocity of the body.
Q13) Which .................... doubling the mass or doubling the velocity?
Sol. Doubling the velocity.
Q14) How.................... 50 kg run so that his kinetic energy be 625 J ?
Sol. 12mv2=K.E
⇒ v=(√625)×2/ 50 =5m/s
Q15) State whether....................................... or both :
(a) A man climbing a hill (b) A flying aeroplane
(c) A bird running on the ground (d) A ceiling fan in the off position (e) A stretched spring lying on the ground.
Sol. (a) Both kinetic energy and potential energy
(b) Both kinetic energy and potential energy
(c) Only kinetic energy (d) Only potential energy
(e) Only potential energy (rather elastic potential energy).
Q.16 Two bodies A and B of equal.................... . What will be the ratio of their potential energies ?
Sol. Ratio=1:2
Q.17 What is the kinetic energy of a........................................ ?
Sol. 1/ 2mv2=1/2×1×22=2J
Q.18 Is potential.................... ?
Sol. Scalar Quantity
Q.19 A load of 100 kg is pulled up by 5 m...................... (g = 9.8 m/s2)
Sol. W=mgh=100×9.8×5=4900J
Q.20 State.................... is true or false. The.................... a body of mass 1 kg kept at a height of 1 m is 1 J.
Sol. False.
Q.21 What........................................ when its height is doubled ?
Sol. Potential energy gets doubled.
Q.22 What kind of energy is.................... ?
(a) A stone kept on roof-top. (b) A running car.
(c) Water stored in the reservoir of a dam.
(d) A compressed spring. (e) A stretched rubber band.
Sol. (a) Potential energy (b) Kinetic Energy
(c) Potential Energy (d) Potential Energy
(e) Potential Energy
Q.23 Fill in the following blanks with suitable words :
(a) Work is measured as a product of.......and.....
(b) The work done on a body moving in a circular path is ..........
(c) 1 joule is the work done when a force of one.....moves an object through a distance of one....in the direction of.....
(d) The ability of a body to do work is called......The ability of a body to do work because of its motion is called......
(e) The sum of the potential and kinetic energies of a body is called......energy.
Sol. (a) Work is measured as a product of force and distance
(b) The work done on a body moving in a circular path is zero
(c) 1 joule is the work done when a force of one Newton moves an object through a distance of one meter in the direction of force
(d) The ability of a body to do work is called energy. The ability of a body to do work because of its motion is called Kinetic Energy
(e) The sum of the potential and kinetic energies of a body is called mechanical energy.
SHORT ANSWER TYPE QUESTIONS:
Q.24 What are the quantities........................................ ? How are they related to work?
Sol. The amount of work done on a body depends
i)Force ii)Displacement.
Work = Force × displacement in the direction of force.If displacement is in a certain direction to the applied force then work done is calculated by force x displacement.
Q.25 Is it possible.................... work done is zero? Explain giving one example.
Sol. i)Displacement is 0 or the initial point and final point are the same.For e.g,When a car is moving on a road, there will be a frictional force applied by the road on the. Using Newton’s third law we can say that, for every action, there is an equal and opposite reaction. Thus the force applied by the road on the car will be equal and opposite to the force applied by the car on the road. Since, there is no displacement of the road, there will be no work done on the road.
ii) Displacement is in perpendicular direction to force applied. For e.g,No work is done by the porter in carrying the load. As the porter carries the load by lifting it upwards and the moving forward it is obvious the angle between the force applied by the porter and the displacement is 90o (i.e.,if Cos θ is zero or θ = Π/2).
Q.26 A boy throws a rubber ball vertically upwards. What type of work, positive or negative is done:
(a) By the force.................... ?
(b) By the.................... earth?
Sol) A boy throws a rubber ball vertically upwards. Work done by the two forces
a) Work done by the force applied by the boy is positive because the displacement of the ball is in the direction of the force applied.
b) Work done by the gravitational force of earth is negative because the displacement of the ball is opposite to the direction of the gravitational force.
Q.27 Write the........................................ at an angle to the direction of for the meaning of each symbol used.
Sol. W = F × d × cos θ where F is force, d is displacement, θ is the angle between F and d.
Q.28 How does the kinetic energy.................... its (i) speed, and (ii) mass?
Sol. Kinetic energy of a moving body depend on its
(i) Speed= Kinetic energy of a moving body is directly proportional to the square of speed of the moving body. K.E∝ v2
(ii) Mass= Kinetic energy of a moving body is directly proportional to the mass of the moving body. K.E∝ m
Q.29 Give one example ....................
(a) positive work (b) negative work, and
(c) zero work
Ans:-
(a) A man carrying a suitcase in his hand climbs up the stairs. The work done by the man is positive because the displacement takes place in the direction of force.
(b)When a body is being lifted in the upward direction. In this case, the force of gravity is acting in the downward direction and the displacement of the body is in the upward direction. As the angle between the force and displacement is 180o, the work done by the gravitational force on the body is negative. Work done by a force is negative if the applied force has a component in a direction opposite to the displacement.
(c)A boy pushing against a tree it does not move, d=0.Then work done is zero.
Q.30 A ball of mass 200 g falls from a height of 5 meters..................... (g = 9.8 m/s2).
Sol) Mass of the ball= 200g= 0.2Kg
Height from which ball is dropped=5m
Initial velocity of the ball,(u)=0m/s
Acceleration due to gravity(g)= 9.8m/s
Therefore, final velocity(v) of the ball is
v2=u2+ 2as
By substituting the above given values, we get
v2 = 0+ 2(9.8) (5)
v2 =98
To find kinetic energy, we use
K.E= 1/2mv2
K.E= ½ x 0.2 x 98
K.E= 9.8 J Thus, kinetic energy =9.8 J
Q.31 Find.................... mass 100 g having a kinetic energy of 20 J.
Sol) K.E. = 20 J(given); Mass m = 100 g = 0.1 kg(given) and Momentum(p) = ?K.E. = ½ mv2
20= ½ x 0.1 x v2
40 = 0.1 x v2 400 = v2
20=v. Therefore, v=20m/s
Momentum(p)= mv
p= 0.1 kg x 20 m/s
p = √(4 (kg.m/s)2) = 2 kg.m/s
Therefore, momentum=2kgm/s
Q.32 Two objects........................................ of 2 m/s and 6 m/s respectively. Calculate the ratio of their kinetic energies.
Sol) Let the velocity of the first body(v1)=2m/s(given)
And let the mass of the first body(m1)= ‘m’kg
Similarly, let the velocity of the second body(v2)=6m/s(given)
And let the mass of the second body(m2)= ‘m’ kg
K.E= 1/2mv2
Therefore K.E of the first body= ½ x m x (2)2
K.E of the second body= ½ x m x (6)2
Ratio of their kinetic energies=½ x m x (2)2/ ½ x m x (6)2
= 4/36
= 1/9
Therefore,
Ratio of the kinetic energy of body 1 : ratio of the kinetic energy of body 2 =1:9
Q.33 A body of 2 kg falls from rest. What.................... the fall at the end 2 s? (Assume ‘g’ =10m/s2)
Sol) Given,
Mass of the body=2kg
Initial velocity(u)=0 m/s
Time(t)=2s
Acceleration due to gravity(g)=10m/s2
Final velocity(v)=?
v=u + at
v= 0 + 10 x 2
v=20 m/s
K.E =1/2 mv2
K.E= ½ x 2 x (20)2
K.E=400J
Kinetic energy of the body during the fall at the end 2 s =400 J
Q.34 On a level road, a scooterist applies.................... 10 m/s to 5 m/s. If the mass of scooterist and the scooter be 150 kg, calculate.................... brakes. (Neglect air resistance and friction)
Sol)Mass of scooterist and the scooter = 150 kg
Initial velocity(v1)=10 m/s
Final velocity(v2)=5 m/s
K.E= ½ mv2
Initial kinetic energy= 1/2 mv12
= ½ x 150 x (10)2
= 7500J
Final kinetic energy=1/2 mv22
= ½ x 150 x (5)2
= 1875 J
Work done by the brakes= change in kinetic energy
= (K.E)2 - (K.E)1
= 1875-7500
=-5625J
The negative sign indicates that the force applied by the breaks is opposite to the direction of motion of the body.
Q.35 A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed.................... when it reaches the ground? ( g = 10 m/ s2)
Sol) Mass of the rock= 10kg
Height from which the rock is dropped= 5m
Initial velocity(u)=0m/s
Let the final velocity= v m/s
Acceleration due to gravity (g) =10m/s2
Therefore, the velocity of the body when it reaches ground can be found out using
v2−u2 = 2as
v2=2×10×5=100
K.E = 1/2 mv2=1/2×10(10)2=500 J
Kinetic energy of the rock=500J
Q.36 Calculate the.................... 1000 kg when its speed is reduced from 20 m/s to 10 m/s?
Sol) Mass of the car= 1000kg
Initial velocity(v1)=20m/s
Final velocity(v2)= 10m/s
Initial kinetic energy= 1/2 mv12
= ½ x 1000 x (20)2
= 200,000J
Final kinetic energy=1/2 mv22
= ½ x 1000 x (10)2
= 50,000 J
Work done by the brakes= change in kinetic energy
= (K.E)2 - (K.E)1
=50,000 – 200,000
=-150000J
= -150KJ
The negative sign indicates that the force applied by the breaks is opposite to the direction of motion of the body.
Q.37 A body of mass 100 kg is lifted up by 10 m. Find :
(i) The amount of work done.
(ii) Potential energy of the body at that height (value of g = 10 m/s2)
Sol. (i) Work done = mgh
= 100 × 10 × 10 = 10,000 J.
(ii) Potential energy = Work done(At any point above the surface of the ground, the work done in raising the body is numerically equal to the potential energy)
Potential energy = 10,000 J.
Potential energy= 10KJ
Q.38 A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate.................... potential energy does he gain? (g = 9.8 m/s2)
Sol.
Given, mass=50kg, g=9.8 m/s2, height=100m
Potential energy = Work done = 49KJ
(At any point above the surface of the ground, the work done in raising the body is numerically equal to the potential energy)
Potential energy =49KJ
Q.39 When is the work done.................... :
(a) positive (b) negative (c) zero
Sol.
Work done is positive, when a force acts in the direction of motion of the body.
Work done is negative, when a force acts opposite to the direction of motion of the body
Work done is zero when a force acts at right angles to the direction of motion of the body
Q.40 To.................... 150 kg be lifted, so that.................... may become 7350 joule? (g- = 9.8 m/s2)
Sol. Given, mass=150 kg, g = 9.8 m/s2 , P.E= 7350 J
Therefore, h=P.E/mg=7350/150×9.8
=7350/1470
= 5m.
Q.41 A body of mass 2 kg.................... velocity of 20 m/s. What.................... energy at the end of 2s ? (Assume g = 10 m/s).
Sol.
Given, Mass=2kg, Initial velocity (u) = 20 m/s, time (t) =2s, g =10 m/s.
Let the height at the end of 2s= d
d=ut+1/2at2
d=20 x 2 + ½ x (-10) x (2)2
d= 40- 20
d=20m
(g is taken as negative because it is acting in the direction opposite to the direction of motion)
P.E = mgh = 2 × 10 × 20 =400 J.
Q.42 How much work is done when a force of I N moves a body through a distance of 1 m in its own direction
Sol
Work done = F × d = 1 × 1 = 1 J.
Q.43 A car is being driven by a force of 2.5 x 1010 .................... ,it takes 2 minutes to reach certain place. Calculate the work done.
Sol) Given, Force = 2.5×1010N
Speed of the car (u) =5m/s
Time taken( t)=2×60s=120s
Distance= Speed x time
= 5 x 120
Work done = Force × distance
= 2.5×1010 × 600
Work done = 1.5×1013J.
Q.44 Explain by an example.................... in motion.
Sol.) The water in a tank on the roof of a building possesses potential energy due to its position (height) above the ground.
Q.45 (a) On what factors.................... potential energy of a body depend?
(b) Give.................... possessing: (i) Kinetic Energy, and (ii) Potential Energy.
Sol.
(a) The gravitational potential energy depends on
(i) Mass of body. (ii) Vertical distance or height
(iii) Acceleration due to gravity
(b) A stationary stone lying at the top of a hill has only potential energy.
When the stone reaches the bottom of the hill, it has only kinetic energy.
Q.46 Give two examples.................... as well as potential energy.
Sol)A flying aeroplane has kinetic energy as well as potential energy.
A man climbing a hill has kinetic energy as well as potential energy.
Q.47 How much is.................... in climbing a tree 5 m tall? (g = 10 m/s2)
Sol) Work done in climbing=2500J
Acceleration due to gravity(g)=10m/s2
Height above the ground= 5m
= 2500/ 10×5=50Kg.
Mass of the man=50 kg
Q.48 If the work done.................... distance of 20 cm is 24.2 J, what is the magnitude of the force?
Sol) Work done in moving a body= 24.2J
Distance travelled by the body=20cm=0.2m
F=w/ d=24.2/ 0.2=121N.
Magnitude of the force =121N
Q.49 A boy.................... jump of 1.5 m.
(i) what is.................... at the highest point?
(ii) What is.................... the highest point? (g = 10 m/s2).
Sol.
(i) K.E = 0 at highest point because velocity of the high jumper at the highest point will be 0
(ii) P.E = mgh = 40 × 10 × 1.5 = 600 J.
Q.50 What type of energy is possessed:
(a) .................... rubber strings of a catapult?
(b) By the piece of stone which is.................... of catapult?
Sol. (a) Potential Energy. (b) Kinetic Energy
Q.51 A weightlifter.................... a height of 2 metres. If g = 9.8 m/s2, calculate :
(a) Potential energy acquired by the weights.
(b) Work done by the weightlifter.
Sol. (a) P.E = mgh = 200 × 9.8 × 2 =3920 J.
(b) Work done = Potential energy gained = 3920 J.
Q.52 (a) Define the term 'work'. Write the formula for the work done on a body when a force acts on the........................................
(b) A person of mass 50 kg climbs.................... work done. (g = 9.8 m s−2)
Sol. (a)Work is done whenever a force acts on a body and the body moves in the direction of force.
When force is exerted on an object and object is displaced, work is said to be done.
Work = Force x Displacement
Or, W = F x s
Where, W is work
‘F’ is force and ‘s’ is displacement.
(b) Work done = mgh = 50 × 9.8 × 72 = 35280 J.
Q.53 (a) When.................... work is done ? Write the formula.................... moving up against gravity. Give the meaning of each symbol which occurs in it.
(b) How much.................... through a distance of 10 cm in the direction of force ?
Sol.
Work is said to be done whenever the following conditions are satisfied-
i)A force acts on a body
ii)There is displacement of the body that is caused by the applied force along the direction of the applied force,.i.e., object is displaced
Formula for the work done by a body in moving up against gravity
Work done = mgh, where m is mass , g is acceleration due to gravity and h is height
(b) Given, F=2N, distance travelled by the body= 10cm=0.1m
W = F × d
= 2N×0.1m = 0.2J.
Work done = 0.2J
Q.54 (a) What.................... when the displacement of a body is at right angles.................... Explain your answer.
(b) A force of 50 N acts.................... 4 m on a horizontal surface. Calculate the work done.................... 60° to the horizontal surface.
Sol
(a) Work done is zero , since there is no displacement in the direction of force
WE know,
W=F S cos θ, where W=work done, F=force, S= displacement, θ= angle between force and distance
When force is perpendicular to the direction of motion, Cos 900
(b)Given, F=50N, S=4m, θ= 600
= 50 × 4 × cos60° = 100 J [As cos60∘=12]
Q.55 (a) Define the term 'energy' of a body. What is the SI unit of energy.
(b) What are the.................... energy ?
(c) Two bodies having.................... speeds of v and 2v respectively.
Find the ratio of their kinetic energies.
Sol) (a) Energy is the capacity to do work.The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule (J). 1 J is the energy required to do 1 joule of work. Sometimes a larger unit of energy called kilo joule (kJ) is used. 1 kJ equals 1000 J.
(b) Various forms of energy are kinetic energy, potential energy, chemical energy, heat energy, electrical energy, light Energy , heat Energy , Nuclear energy
(c) Let the masses of the two bodies(m1) and (m2) be m kg
Let the velocity of the first body be v m/s
And the velocity of the second body be 2v m/s
Ratio of K.E = (KE) of body 1/ KE of body 2
=( ½ mv2)/ (1/2m(2v)2)2
=v2/(2v)2= (1)2/ (2)2
= 1:4( K.E∝v2)
Ratio of their kinetic energies =1:4
Q.56 (a) What.................... kinetic energy of a body?
(b) A body is.................... decreasing. What happens to its kinetic energy as its velocity becomes zero?
(c) A horse and a dog are running.................... horse is ten times that of the dog, what is the ratio of their kinetic energies?
Sol. (a) Energy possessed by a body due to its motion is called its kinetic energy.Kinetic energy is directly proportional to the mass and square of the velocity of the moving body. K.E= 1/2mv2
(b) When a body is thrown vertically upwards against the force of gravity, its Kinetic energy goes on decreasing as its velocity decreases. At maximum height, the energy becomes zero as velocity becomes zero and all the energy in the body is transformed into potential energy. As KE decrease PE increase.
(c) As K.E∝ mass
Ratio of KE = K.E of horse/K.E of dog
= 10/1=10
Ratio of their kinetic energies= 10:1
Q.57 (a) Explain by an example.................... . Write down the expression for gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.
(b) What is the difference between.................... ?
(c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s..................... State your answer giving proper units.
Sol) (a) Potential energy is defined as the energy possessed by a body due to its position. For example; when a stone is kept at a height, it possesses some energy because of its height. Because of this potential energy, object kept at a height falls over the ground.
Expression for Potential Energy:
P.E = mgh, where m is mass , g is acceleration due to gravity and ‘h’ is the height .
(b)Potential Energy---Potential energy is the energy possessed by a body due to its position or its configuration Kinetic energy is the energy possessed by the body due to its motion .
Kinetic Energy -----It is independent of the speed of the body Kinetic energy is directly proportional to the square of the velocity of the moving body .
E.g.,A rock sitting at the edge of a cliff has potential energy E.g.,A rock falling from a cliff.
(c) Initial K.E = 1/2mv2=1/2×0.5× (5)2
Final K.E = 1/2mv2=1/2×0.5× (3)2
Work done=Change in K.E
= 2.25- 6.25
= - 4 J.
Q.58 (a) What is the difference between gravitational potential energy and elastic potential energy ? Give one example of a body ....................
(b) If 784 J of work was done in lifting a 20 kg mass,........................................ lifted. (g = 9.8 m/s2)
Sol) Energy of a body due to its position above the ground is called gravitational potential energy. Energy of a body due to a change in its shape and size is called elastic potential energy.Elastic property of the body is responsible for this energy.Elastic potential energy is associated with the state of compression or extension of an object.
For example, a wound –up circular spring possess elastic potential energy which drives a wound-up toy.
b) Mass of a body= 20kg
Acceleration due to gravity (g) =9.8m/s2
Work done= 784 J
W= m x g x h
h= W/m x g
h= 784/ 20 x 9.8
h=784/196
h=4m
Therefore, the height through which the body was lifted=4m
Questions Based on High Order Thinking Skills(HOTS)
Q.71 A boy tries to push a truck parked on the roadside. The truck does not move at all. Another boy pushes a bicycle. The bicycle moves through a certain distance. In which case was the work done more: on the truck or on the bicycle? Give a reason to support your answer.
Sol) The boy pushing the truck does no work as displacement is zero(no motion of the truck). But, when the boy pushes the bicycle, work done is more as the force applied by him causes displacement (motion of the bicycle).
Q.72 The work done by a force acting obliquely is given by the formula W = F cosθ x s. What will happen to the work.................... is increased gradually? Will it increase, decrease or remain constant?
Sol. On increasing the value of angle between force and distance, the value of cos θ decreases to zero when θ=90∘Therefore, work done decrease as angle increase, work done is 0 when θ=90∘ and work done is negative when angle further increases.
cos θ reduces further to –1.
Therefore, work done decrease as angle increase, work done is 0 when θ=90∘ and work done is negative when angle further increases.
Q.73 What should be the angle.................... of a body so that the work done is zero?
Sol) When the force is perpendicular to the direction of motion,work done will be zero. i.e., the angle should be 90∘ between force and displacement, for the work to be zero.
Q.74 In which of.................... when the angle between the direction of force and direction of motion is 0° or 90°?
Sol. When angle is 0∘
Q.75 How much work is done by the.................... moving around it in a circular path? Give reason for your answer.
Sol) Work done is zero, as force and displacement are at 90∘ to each other.
Q.76 A man is.................... of a hill at a height of 1200 meters. The man weighs 800 N and the package weighs 200 N. If g =10 m/S2.
(i) How much work does man do against gravity?
(ii) What is the potential energy of the package at A if it is assumed to be zero at B?
Sol. (i) Weight of the man= 800N
Weight of the package= 200N
Total weight= 800 + 200= 1000N
Height of the hill= 1200m
g=10m/s2
(ii)Given, Weight of the package(mg)= 200N, h=1200m
P.E = m x g x h
= 200× 1200
=2.40000J
=2.4 KJ
Q.77 When a ball is thrown vertically upwards,.................... What happens to its potential energy as its velocity becomes zero?
Sol) The potential energy keeps increasing as velocity keeps decreasing. It is maximum when velocity is zero.
Q.78 A man X goes to the top of a building by a vertical spiral staircase. Another man Y of the same mass.................... ladder. Which of the two does more work against gravity and why?
Sol. Work done against gravity depends on the vertical distance through which the body is lifted and is independent of the path travelled to achieve the height. Therefore, we can say that both men did the same work as the vertical distance travelled by them is the same.
Q.79 When a ball is thrown inside a moving bus, does its kinetic.................... the bus? Explain.
Sol. Yes. The kinetic energy of the ball thrown inside a moving bus is dependent on the speed of the moving bus. This is because the speed of the bus adds up to the speed with which the ball is thrown inside the moving bus.
For example, If the bus is moving at a speed of 30 km.hr and a person inside the bus throws a ball with a speed of 40 km/hr in the same direction as that of the motion of the bus. Then, for a person inside the bus, the K.E= ½ x m x (30)2. But, for a person who is static and is outside the bus, the K.E of the ball=1/2 x m x (30+ 40)2. Thus, the motion of the bus affects the K.E of the ball.
Q.80 A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy ? The bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to kinetic energy originally in the bullet?
Sol. Given, mass= 15g = 0.015kg; Initial velocity= 400m/s, distance, s=2cm=0.02m
K.E(initial)=1/2mv2
= ½ x 0.015 x (400)2
K.E(initial) = 1200J
As final velocity = 0.
Therefore, K.E(final)=0 ( Where F is the average force )
F=1200/ 2×10−2= -6×104N.
Negative sign shows that force is acting opposite to the direction of motion of the bullet. The kinetic energy originally in the bullet is eventually converted to heat energy by friction.
Very Short Answer Type Questions:
Q1) Name the.................... of energy.
Sol. KWh
Q2) Define kilowatt-hour.
Sol) 1 Kilowatt hour is the energy consumed when 1000 Watt is consumed for 60 minutes.
Q3) Name two units of.................... than watt
Sol) Kilowatt (KW) and Megawatt (MW)
Q4) Define the term 'watt'.
Sol. One Watt is defined as the power when one Joule of work is done in 1 s.
Q5) How many.................... one horse power?
Sol. 746 watt
Q6) Name the.................... whose unit is watt.
Sol. Power
Q7)What is the power of.................... the rate of one joule per second ?
Sol. 1 W
Q.8) A body does 1200 joules of work in 2 minutes. Calculate its power.
Sol. work =1200 J, time= 2 min= 2x 60 sec= 120 sec
Power = work done/time taken
= 1200/120
=10 W
Q9) How many joul.................... one kilowatt-hour ?
Sol. 3.6×106J.
Q10) Name the quantity whose unit is :
(a) kilowatt (b) kilowatt-hour
Sol. (a) Power (b) Energy.
Q.11) What is the common name of '1 kWh' of electrical energy ?
Sol. Unit of electricity
Q12) A cell converts.................... form. Name the two forms.
Sol. A cell converts Chemical energy into Electrical energy.
Q13) Name the device which.................... .
Sol. Electric motor
Q14) Name the devices or machines which convert :
(a) Mechanical energy into Electrical Energy.
(b) Chemical energy into Electrical Energy.
(c) Electrical energy into Heat Energy.
(d) Light energy into Electrical Energy.
(e) Electrical energy into Light Energy.
Sol)
(a) Electric generator (b) Battery
(c) Electric iron (d) Solar cell. (e) Electric bulb
Q15)Name the devices or machines which convert :
(i) Electrical energy into sound energy.
(ii) Heat energy into kinetic energy (or mechanical energy).
(iii) Chemical energy into kinetic energy (or mechanical energy).
(iv) Chemical energy into heat energy.
(v) Light energy into heat energy.
Sol) (i) Speaker in radio and television. (ii) Steam engin
(iii) Car engine. (iv) Gas stove (v) Solar water heater.
Q16) Fill in the following blanks with suitable words :
(a) Power is the rate of doing.......
(b) 1 watt is a rate of working of one....per.....
(c) The electricity meter installed in our home measures electric energy in the units of......
(d) The principle of......of energy says that energy can be......from one form to another, but it cannot be......or......
(e) When a ball is thrown upwards......energy is transformed into.......energy.
Sol. (a) Power is the rate of doing work
(b) 1 watt is a rate of working of one joule per second
(c) The electricity meter installed in our home measures electric energy in the units of kWh
(d) The principle of conservation of energy says that energy can be transformed from one form to another, but it cannot be created or
(e) When a ball is thrown up wards kinetic energy is transformed into potential energy.
Short Answer Type Questions:
Q17) A trolley is pushed a.................... in 1 minute. Calculate the power developed.
Sol. Force= 400N, distance travelled= 60m, t= 1 min=60 s
Work done= Force x distance
Work done= 400 x 60
Work done= 24000 J
Power = Work done/Time
Power= 24000/60
= 400 W.
Q18) What kind of energy.................... power station?
Sol. Potential energy of water → K.E of water → K.E of turbines → Electrical Energy.
Q19) What kind of.................... coal-based thermal power station?
Sol. Chemical energy → Heat energy → Kinetic energy of steam → K.E of turbines → Electrical Energy.
Q20) A man weighing.................... 4 m high in 5 seconds. What is the power?
Sol) Given, weight of the man= 500N
Weight of the load=100N
Total weight= 500 + 100= 600N
Height(vertical distance)= 4m
Work done= total weight x vertical distance
Work done= 600 x 4
Work done= 2400 J
Power = Work done/time
=2400/ 5=480W.
Q21) The power output of an engine is 3 KW. How much.................... in 20 s?
Sol) Given, Power= 3KW=3000W; t=20s
= 3000 × 20 = 60000 J = 60 KJ.
Work done =60KJ
Q.22) An electric.................... in 5 minutes. Calculate its power rating.
Sol. Energy= 600KJ= 600000J; time= 5 min= 5 x 60= 300s
Power = Energy/ Time=600000/300
Power= 2000W
Power of electric heater = 2000W
Q23)How much.................... does a 100 Watt lamp consume :
(a) in 1 second ? (b) in 1 minute ?
Sol) Given, power (P)= 100 watt
t=1 s
Energy = P × T = 100 × 1 = 100 J
(b) t=1min= 60 sec
Energy = 100 × 60 = 6000 J.
Q24) Five electric fans of 120 watts.................... the electrical energy consumed in kilowatt-hours.
Sol. Given, power= 120 watts=0.12 kw; t=4 hr
Energy consumed by one fan = 0.12 × 4 = 0.48 KWh.
Therefore, energy consumed by 5 fans= 0.48 x 5=2.4 kwh
Q25) Describe the energy changes which take place in a radio.
Sol) Electrical Energy →Kinetic Energy→ Sound Energy.
Radio converts electrical energy into kinetic energy and finally into sound energy
Q26) Write the energy.................... in an electric bulb (or electric lamp).
Sol) Electrical Energy → Heat Energy → Light Energy.
Q 27) Name five appliances or machines which use an electric motor.
Sol) Electrical fans, washing machine, refrigerator, grinder, electric iron.
Q 28)A bulb lights up.................... State the energy change which takes place:
(i) in the battery. (ii) in the bulb.
Sol)
(i) Chemical energy → Electrical energy.
(ii) Electrical Energy → Heat Energy → Light Energy
Q29)The hanging bob of a simple pendulum.................... released. It swings towards center position A and then to the other extreme position C.In which position do the bob have:
(i) maximum potential energy?
(ii) Maximum kinetic energy? Give reasons for your answer.
Sol. (i) Maximum P.E at the extreme position B and C because at these positions, the bob is at maximum height from the ground. K.E at these points is zero.
(ii) Maximum K.E at A because at these position, the bob has maximum velocity. Potential energy at this point is zero.
Q30) A car of weight 20000 N.................... 8 m/s, gaining a height of 120 m in 100 s. Calculate:
(a) Work done by the car.
(b) Power of engine of car.
Sol) (a) Weight of the car(W)= 20000 N
Height=120 m
Time taken (t)= 100 s
Work done by car = F × d =20,000 × 120 = 2400000 KJ or 2.4 x 106J
(b) Power = work done/ time=2400000/ 100=24 KW.
Long Answer Type Questions:
Q31) (a) What do you.................... term "transformation of energy”? Explain with an example.
(b) Explain the.................... following cases :
(i) A ball thrown upwards.
(ii) A stone dropped from the roof of a building.
Sol.
(i) K.E → P.E
(ii) P.E → K.E
Q.32
(a) State.................... of energy with an example.
(b) Explain how, the total energy.................... remains conserved. Illustrate your answer with the help of a labeled diagram.
Sol.
Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.Law of conservation of Energy states that the total energy of a system remains unchanged before and after transformation.
For example- a) A rock on a cliff. The rock has potential energy, when it's pushed off from a height, the potential energy is changed into kinetic energy.
The sum of potential energy and kinetic energy remains constant at every point of the falling of object.
mgh + ½ mv2= constant at every point
The sum of potential energy and kinetic energy is the total mechanical energy of the rock falling from a height(cliff).
During free fall of the object, the potential energy starts decreasing and converting into kinetic energy with decrease of height from the ground
(b) A swinging pendulum is a perfect example to show the conservation of energy. It shows the transformation of potential energy into kinetic energy and kinetic energy back into potential energy without any energy loss. In a pendulum, the law establishes that, when the ball is at its highest point, all the energy is potential energy and there is zero kinetic energy. At the ball's lowest point, all the energy in the ball is kinetic and there is zero potential energy. The total energy of the ball is the sum of the potential energy and kinetic energy.
Initially, the bob of the pendulum is at the mean position(B). When we draw the pendulum bob to one side(Extreme position A), we raise the bob to a little height end give it potential energy. This is the energy transferred by work done by hand. As at the extreme position, the bob has only PE, it tends to move down.The P.E decreases and K.E increases. At the lowest(mean) position, the bob has got K.E. Due to this it moves to the other side. Now, its K.E decreases and P.E increases. At the extreme positions A and C, all energy is in the form of potential energy and therefore it tends to move down. Thus the bob oscillates. At all other intermediate positions, energy of the pendulum is partly potential and partly kinetic. But, the total energy of the pendulum remains conserved.
Q33) (a) What is the meaning of the symbol KWh? What quantity does it represent?
(b) How much electric energy in KWh.................... of 1000 watts when it is switched on for 60 minutes?
Sol) (a) KWh means kilowalt hour. It is the commercial unit of measurement of electrical energy. 1 kWh = 3.6·106 J. It can be defined as the amount of electrical energy consumed when an electrical appliance of 1KW power runs for one hour.
(b) Power= 1000 watts(1KW); time= 60 min=1 hr
Energy = P × t = 1000 × 1h = 1000 Wh = 1 KWh.
Q34) (a) Derive the relation between commercial unit of energy (KWh) and SI unit of energy (joule).
(b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joules ?
Sol) (a) 1kWh = 1000 W × 1h.
1 watt= 1 joule/1 sec
= 1000 × 60 × 60
= 3.6×106J
Q34)(a) Derive the relation between.................... (KWh) and SI unit of energy (joule).
(b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joules ?
Sol)(a) 1kWh = 1000 W × 1h.
1 watt= 1 joule/1 sec
= 1000 × 60 × 6 = 3.6×106J
(b) Gn, total units consumed= 650 units
1 unit of electricity = 1 KWh(i.e., KWh is also known as unit)
650 units = 650 KWh. = 650 × 3.6 × 106J
= 2.34 × 109J
Q35)(a) Define power. Give the SI unit of power.
(b) A boy weighing 40 kg carries.................... 15 m high in 25 seconds. Calculate the power. (g = 10m/s2)
Sol) (a) Power is the rate of doing work . It is equivalent to an amount of energy consumed per unit time. In the SI system, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the steam engine.
1 Watt is the power of an object which does work at the rate of 1 Joule per second.
(b) Gn, mass of boy= 40Kg
Mass of box=20 kg
Total mass= 40+20=60kg
g = 10m/s2
h= 15m
t=25 sec
Power =mgh/t =60×10×15/ 25=360 W
Power= 360W
Very Short Answer Type Questions
Q1) How much work.................... m is raised to a height h above the ground ?
Sol. Work done = m × g × h , where g is acceleration due to gravity.
Q2) State.................... of work.
Sol. Joule.
Q3) Is work.................... quantity ?
Sol. Scalar.
Q4) Define 1 Joule ....................
Sol. 1 Joule of work is done when 1 N force moves a body by 1 m.
Q.5 What is the condition.................... on a body ?
Sol. The force should produce motion in the body.
Q6) Is energy.................... quantity ?
Sol. No.
Q7) What a.................... of (a) work, and (b) energy ?
Sol. (a) Joule (b) Joule
Q8) What is the work.................... along a friction less surface?
Sol. Zero.
Q9) By how much.................... if its speed is doubled?
Sol. The kinetic energy of a body will become four times.
Q10) Write an expression ....................mass m moving with a velocity v.
Sol. 1/2 mv2
Q11) If the speed of a body is halved,.................... its kinetic energy ?
Sol. It becomes one-fourth
Q12) On .................... kinetic energy of a body depend ?
Sol. It depends on mass and velocity of the body.
Q13) Which .................... doubling the mass or doubling the velocity?
Sol. Doubling the velocity.
Q14) How.................... 50 kg run so that his kinetic energy be 625 J ?
Sol. 12mv2=K.E
⇒ v=(√625)×2/ 50 =5m/s
Q15) State whether....................................... or both :
(a) A man climbing a hill (b) A flying aeroplane
(c) A bird running on the ground (d) A ceiling fan in the off position (e) A stretched spring lying on the ground.
Sol. (a) Both kinetic energy and potential energy
(b) Both kinetic energy and potential energy
(c) Only kinetic energy (d) Only potential energy
(e) Only potential energy (rather elastic potential energy).
Q.16 Two bodies A and B of equal.................... . What will be the ratio of their potential energies ?
Sol. Ratio=1:2
Q.17 What is the kinetic energy of a........................................ ?
Sol. 1/ 2mv2=1/2×1×22=2J
Q.18 Is potential.................... ?
Sol. Scalar Quantity
Q.19 A load of 100 kg is pulled up by 5 m...................... (g = 9.8 m/s2)
Sol. W=mgh=100×9.8×5=4900J
Q.20 State.................... is true or false. The.................... a body of mass 1 kg kept at a height of 1 m is 1 J.
Sol. False.
Q.21 What........................................ when its height is doubled ?
Sol. Potential energy gets doubled.
Q.22 What kind of energy is.................... ?
(a) A stone kept on roof-top. (b) A running car.
(c) Water stored in the reservoir of a dam.
(d) A compressed spring. (e) A stretched rubber band.
Sol. (a) Potential energy (b) Kinetic Energy
(c) Potential Energy (d) Potential Energy
(e) Potential Energy
Q.23 Fill in the following blanks with suitable words :
(a) Work is measured as a product of.......and.....
(b) The work done on a body moving in a circular path is ..........
(c) 1 joule is the work done when a force of one.....moves an object through a distance of one....in the direction of.....
(d) The ability of a body to do work is called......The ability of a body to do work because of its motion is called......
(e) The sum of the potential and kinetic energies of a body is called......energy.
Sol. (a) Work is measured as a product of force and distance
(b) The work done on a body moving in a circular path is zero
(c) 1 joule is the work done when a force of one Newton moves an object through a distance of one meter in the direction of force
(d) The ability of a body to do work is called energy. The ability of a body to do work because of its motion is called Kinetic Energy
(e) The sum of the potential and kinetic energies of a body is called mechanical energy.
SHORT ANSWER TYPE QUESTIONS:
Q.24 What are the quantities........................................ ? How are they related to work?
Sol. The amount of work done on a body depends
i)Force ii)Displacement.
Work = Force × displacement in the direction of force.If displacement is in a certain direction to the applied force then work done is calculated by force x displacement.
Q.25 Is it possible.................... work done is zero? Explain giving one example.
Sol. i)Displacement is 0 or the initial point and final point are the same.For e.g,When a car is moving on a road, there will be a frictional force applied by the road on the. Using Newton’s third law we can say that, for every action, there is an equal and opposite reaction. Thus the force applied by the road on the car will be equal and opposite to the force applied by the car on the road. Since, there is no displacement of the road, there will be no work done on the road.
ii) Displacement is in perpendicular direction to force applied. For e.g,No work is done by the porter in carrying the load. As the porter carries the load by lifting it upwards and the moving forward it is obvious the angle between the force applied by the porter and the displacement is 90o (i.e.,if Cos θ is zero or θ = Π/2).
Q.26 A boy throws a rubber ball vertically upwards. What type of work, positive or negative is done:
(a) By the force.................... ?
(b) By the.................... earth?
Sol) A boy throws a rubber ball vertically upwards. Work done by the two forces
a) Work done by the force applied by the boy is positive because the displacement of the ball is in the direction of the force applied.
b) Work done by the gravitational force of earth is negative because the displacement of the ball is opposite to the direction of the gravitational force.
Q.27 Write the........................................ at an angle to the direction of for the meaning of each symbol used.
Sol. W = F × d × cos θ where F is force, d is displacement, θ is the angle between F and d.
Q.28 How does the kinetic energy.................... its (i) speed, and (ii) mass?
Sol. Kinetic energy of a moving body depend on its
(i) Speed= Kinetic energy of a moving body is directly proportional to the square of speed of the moving body. K.E∝ v2
(ii) Mass= Kinetic energy of a moving body is directly proportional to the mass of the moving body. K.E∝ m
Q.29 Give one example ....................
(a) positive work (b) negative work, and
(c) zero work
Ans:-
(a) A man carrying a suitcase in his hand climbs up the stairs. The work done by the man is positive because the displacement takes place in the direction of force.
(b)When a body is being lifted in the upward direction. In this case, the force of gravity is acting in the downward direction and the displacement of the body is in the upward direction. As the angle between the force and displacement is 180o, the work done by the gravitational force on the body is negative. Work done by a force is negative if the applied force has a component in a direction opposite to the displacement.
(c)A boy pushing against a tree it does not move, d=0.Then work done is zero.
Q.30 A ball of mass 200 g falls from a height of 5 meters..................... (g = 9.8 m/s2).
Sol) Mass of the ball= 200g= 0.2Kg
Height from which ball is dropped=5m
Initial velocity of the ball,(u)=0m/s
Acceleration due to gravity(g)= 9.8m/s
Therefore, final velocity(v) of the ball is
v2=u2+ 2as
By substituting the above given values, we get
v2 = 0+ 2(9.8) (5)
v2 =98
To find kinetic energy, we use
K.E= 1/2mv2
K.E= ½ x 0.2 x 98
K.E= 9.8 J Thus, kinetic energy =9.8 J
Q.31 Find.................... mass 100 g having a kinetic energy of 20 J.
Sol) K.E. = 20 J(given); Mass m = 100 g = 0.1 kg(given) and Momentum(p) = ?K.E. = ½ mv2
20= ½ x 0.1 x v2
40 = 0.1 x v2 400 = v2
20=v. Therefore, v=20m/s
Momentum(p)= mv
p= 0.1 kg x 20 m/s
p = √(4 (kg.m/s)2) = 2 kg.m/s
Therefore, momentum=2kgm/s
Q.32 Two objects........................................ of 2 m/s and 6 m/s respectively. Calculate the ratio of their kinetic energies.
Sol) Let the velocity of the first body(v1)=2m/s(given)
And let the mass of the first body(m1)= ‘m’kg
Similarly, let the velocity of the second body(v2)=6m/s(given)
And let the mass of the second body(m2)= ‘m’ kg
K.E= 1/2mv2
Therefore K.E of the first body= ½ x m x (2)2
K.E of the second body= ½ x m x (6)2
Ratio of their kinetic energies=½ x m x (2)2/ ½ x m x (6)2
= 4/36
= 1/9
Therefore,
Ratio of the kinetic energy of body 1 : ratio of the kinetic energy of body 2 =1:9
Q.33 A body of 2 kg falls from rest. What.................... the fall at the end 2 s? (Assume ‘g’ =10m/s2)
Sol) Given,
Mass of the body=2kg
Initial velocity(u)=0 m/s
Time(t)=2s
Acceleration due to gravity(g)=10m/s2
Final velocity(v)=?
v=u + at
v= 0 + 10 x 2
v=20 m/s
K.E =1/2 mv2
K.E= ½ x 2 x (20)2
K.E=400J
Kinetic energy of the body during the fall at the end 2 s =400 J
Q.34 On a level road, a scooterist applies.................... 10 m/s to 5 m/s. If the mass of scooterist and the scooter be 150 kg, calculate.................... brakes. (Neglect air resistance and friction)
Sol)Mass of scooterist and the scooter = 150 kg
Initial velocity(v1)=10 m/s
Final velocity(v2)=5 m/s
K.E= ½ mv2
Initial kinetic energy= 1/2 mv12
= ½ x 150 x (10)2
= 7500J
Final kinetic energy=1/2 mv22
= ½ x 150 x (5)2
= 1875 J
Work done by the brakes= change in kinetic energy
= (K.E)2 - (K.E)1
= 1875-7500
=-5625J
The negative sign indicates that the force applied by the breaks is opposite to the direction of motion of the body.
Q.35 A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed.................... when it reaches the ground? ( g = 10 m/ s2)
Sol) Mass of the rock= 10kg
Height from which the rock is dropped= 5m
Initial velocity(u)=0m/s
Let the final velocity= v m/s
Acceleration due to gravity (g) =10m/s2
Therefore, the velocity of the body when it reaches ground can be found out using
v2−u2 = 2as
v2=2×10×5=100
K.E = 1/2 mv2=1/2×10(10)2=500 J
Kinetic energy of the rock=500J
Q.36 Calculate the.................... 1000 kg when its speed is reduced from 20 m/s to 10 m/s?
Sol) Mass of the car= 1000kg
Initial velocity(v1)=20m/s
Final velocity(v2)= 10m/s
Initial kinetic energy= 1/2 mv12
= ½ x 1000 x (20)2
= 200,000J
Final kinetic energy=1/2 mv22
= ½ x 1000 x (10)2
= 50,000 J
Work done by the brakes= change in kinetic energy
= (K.E)2 - (K.E)1
=50,000 – 200,000
=-150000J
= -150KJ
The negative sign indicates that the force applied by the breaks is opposite to the direction of motion of the body.
Q.37 A body of mass 100 kg is lifted up by 10 m. Find :
(i) The amount of work done.
(ii) Potential energy of the body at that height (value of g = 10 m/s2)
Sol. (i) Work done = mgh
= 100 × 10 × 10 = 10,000 J.
(ii) Potential energy = Work done(At any point above the surface of the ground, the work done in raising the body is numerically equal to the potential energy)
Potential energy = 10,000 J.
Potential energy= 10KJ
Q.38 A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate.................... potential energy does he gain? (g = 9.8 m/s2)
Sol.
Given, mass=50kg, g=9.8 m/s2, height=100m
Potential energy = Work done = 49KJ
(At any point above the surface of the ground, the work done in raising the body is numerically equal to the potential energy)
Potential energy =49KJ
Q.39 When is the work done.................... :
(a) positive (b) negative (c) zero
Sol.
Work done is positive, when a force acts in the direction of motion of the body.
Work done is negative, when a force acts opposite to the direction of motion of the body
Work done is zero when a force acts at right angles to the direction of motion of the body
Q.40 To.................... 150 kg be lifted, so that.................... may become 7350 joule? (g- = 9.8 m/s2)
Sol. Given, mass=150 kg, g = 9.8 m/s2 , P.E= 7350 J
Therefore, h=P.E/mg=7350/150×9.8
=7350/1470
= 5m.
Q.41 A body of mass 2 kg.................... velocity of 20 m/s. What.................... energy at the end of 2s ? (Assume g = 10 m/s).
Sol.
Given, Mass=2kg, Initial velocity (u) = 20 m/s, time (t) =2s, g =10 m/s.
Let the height at the end of 2s= d
d=ut+1/2at2
d=20 x 2 + ½ x (-10) x (2)2
d= 40- 20
d=20m
(g is taken as negative because it is acting in the direction opposite to the direction of motion)
P.E = mgh = 2 × 10 × 20 =400 J.
Q.42 How much work is done when a force of I N moves a body through a distance of 1 m in its own direction
Sol
Work done = F × d = 1 × 1 = 1 J.
Q.43 A car is being driven by a force of 2.5 x 1010 .................... ,it takes 2 minutes to reach certain place. Calculate the work done.
Sol) Given, Force = 2.5×1010N
Speed of the car (u) =5m/s
Time taken( t)=2×60s=120s
Distance= Speed x time
= 5 x 120
Work done = Force × distance
= 2.5×1010 × 600
Work done = 1.5×1013J.
Q.44 Explain by an example.................... in motion.
Sol.) The water in a tank on the roof of a building possesses potential energy due to its position (height) above the ground.
Q.45 (a) On what factors.................... potential energy of a body depend?
(b) Give.................... possessing: (i) Kinetic Energy, and (ii) Potential Energy.
Sol.
(a) The gravitational potential energy depends on
(i) Mass of body. (ii) Vertical distance or height
(iii) Acceleration due to gravity
(b) A stationary stone lying at the top of a hill has only potential energy.
When the stone reaches the bottom of the hill, it has only kinetic energy.
Q.46 Give two examples.................... as well as potential energy.
Sol)A flying aeroplane has kinetic energy as well as potential energy.
A man climbing a hill has kinetic energy as well as potential energy.
Q.47 How much is.................... in climbing a tree 5 m tall? (g = 10 m/s2)
Sol) Work done in climbing=2500J
Acceleration due to gravity(g)=10m/s2
Height above the ground= 5m
= 2500/ 10×5=50Kg.
Mass of the man=50 kg
Q.48 If the work done.................... distance of 20 cm is 24.2 J, what is the magnitude of the force?
Sol) Work done in moving a body= 24.2J
Distance travelled by the body=20cm=0.2m
F=w/ d=24.2/ 0.2=121N.
Magnitude of the force =121N
Q.49 A boy.................... jump of 1.5 m.
(i) what is.................... at the highest point?
(ii) What is.................... the highest point? (g = 10 m/s2).
Sol.
(i) K.E = 0 at highest point because velocity of the high jumper at the highest point will be 0
(ii) P.E = mgh = 40 × 10 × 1.5 = 600 J.
Q.50 What type of energy is possessed:
(a) .................... rubber strings of a catapult?
(b) By the piece of stone which is.................... of catapult?
Sol. (a) Potential Energy. (b) Kinetic Energy
Q.51 A weightlifter.................... a height of 2 metres. If g = 9.8 m/s2, calculate :
(a) Potential energy acquired by the weights.
(b) Work done by the weightlifter.
Sol. (a) P.E = mgh = 200 × 9.8 × 2 =3920 J.
(b) Work done = Potential energy gained = 3920 J.
Q.52 (a) Define the term 'work'. Write the formula for the work done on a body when a force acts on the........................................
(b) A person of mass 50 kg climbs.................... work done. (g = 9.8 m s−2)
Sol. (a)Work is done whenever a force acts on a body and the body moves in the direction of force.
When force is exerted on an object and object is displaced, work is said to be done.
Work = Force x Displacement
Or, W = F x s
Where, W is work
‘F’ is force and ‘s’ is displacement.
(b) Work done = mgh = 50 × 9.8 × 72 = 35280 J.
Q.53 (a) When.................... work is done ? Write the formula.................... moving up against gravity. Give the meaning of each symbol which occurs in it.
(b) How much.................... through a distance of 10 cm in the direction of force ?
Sol.
Work is said to be done whenever the following conditions are satisfied-
i)A force acts on a body
ii)There is displacement of the body that is caused by the applied force along the direction of the applied force,.i.e., object is displaced
Formula for the work done by a body in moving up against gravity
Work done = mgh, where m is mass , g is acceleration due to gravity and h is height
(b) Given, F=2N, distance travelled by the body= 10cm=0.1m
W = F × d
= 2N×0.1m = 0.2J.
Work done = 0.2J
Q.54 (a) What.................... when the displacement of a body is at right angles.................... Explain your answer.
(b) A force of 50 N acts.................... 4 m on a horizontal surface. Calculate the work done.................... 60° to the horizontal surface.
Sol
(a) Work done is zero , since there is no displacement in the direction of force
WE know,
W=F S cos θ, where W=work done, F=force, S= displacement, θ= angle between force and distance
When force is perpendicular to the direction of motion, Cos 900
(b)Given, F=50N, S=4m, θ= 600
= 50 × 4 × cos60° = 100 J [As cos60∘=12]
Q.55 (a) Define the term 'energy' of a body. What is the SI unit of energy.
(b) What are the.................... energy ?
(c) Two bodies having.................... speeds of v and 2v respectively.
Find the ratio of their kinetic energies.
Sol) (a) Energy is the capacity to do work.The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule (J). 1 J is the energy required to do 1 joule of work. Sometimes a larger unit of energy called kilo joule (kJ) is used. 1 kJ equals 1000 J.
(b) Various forms of energy are kinetic energy, potential energy, chemical energy, heat energy, electrical energy, light Energy , heat Energy , Nuclear energy
(c) Let the masses of the two bodies(m1) and (m2) be m kg
Let the velocity of the first body be v m/s
And the velocity of the second body be 2v m/s
Ratio of K.E = (KE) of body 1/ KE of body 2
=( ½ mv2)/ (1/2m(2v)2)2
=v2/(2v)2= (1)2/ (2)2
= 1:4( K.E∝v2)
Ratio of their kinetic energies =1:4
Q.56 (a) What.................... kinetic energy of a body?
(b) A body is.................... decreasing. What happens to its kinetic energy as its velocity becomes zero?
(c) A horse and a dog are running.................... horse is ten times that of the dog, what is the ratio of their kinetic energies?
Sol. (a) Energy possessed by a body due to its motion is called its kinetic energy.Kinetic energy is directly proportional to the mass and square of the velocity of the moving body. K.E= 1/2mv2
(b) When a body is thrown vertically upwards against the force of gravity, its Kinetic energy goes on decreasing as its velocity decreases. At maximum height, the energy becomes zero as velocity becomes zero and all the energy in the body is transformed into potential energy. As KE decrease PE increase.
(c) As K.E∝ mass
Ratio of KE = K.E of horse/K.E of dog
= 10/1=10
Ratio of their kinetic energies= 10:1
Q.57 (a) Explain by an example.................... . Write down the expression for gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.
(b) What is the difference between.................... ?
(c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s..................... State your answer giving proper units.
Sol) (a) Potential energy is defined as the energy possessed by a body due to its position. For example; when a stone is kept at a height, it possesses some energy because of its height. Because of this potential energy, object kept at a height falls over the ground.
Expression for Potential Energy:
P.E = mgh, where m is mass , g is acceleration due to gravity and ‘h’ is the height .
(b)Potential Energy---Potential energy is the energy possessed by a body due to its position or its configuration Kinetic energy is the energy possessed by the body due to its motion .
Kinetic Energy -----It is independent of the speed of the body Kinetic energy is directly proportional to the square of the velocity of the moving body .
E.g.,A rock sitting at the edge of a cliff has potential energy E.g.,A rock falling from a cliff.
(c) Initial K.E = 1/2mv2=1/2×0.5× (5)2
Final K.E = 1/2mv2=1/2×0.5× (3)2
Work done=Change in K.E
= 2.25- 6.25
= - 4 J.
Q.58 (a) What is the difference between gravitational potential energy and elastic potential energy ? Give one example of a body ....................
(b) If 784 J of work was done in lifting a 20 kg mass,........................................ lifted. (g = 9.8 m/s2)
Sol) Energy of a body due to its position above the ground is called gravitational potential energy. Energy of a body due to a change in its shape and size is called elastic potential energy.Elastic property of the body is responsible for this energy.Elastic potential energy is associated with the state of compression or extension of an object.
For example, a wound –up circular spring possess elastic potential energy which drives a wound-up toy.
b) Mass of a body= 20kg
Acceleration due to gravity (g) =9.8m/s2
Work done= 784 J
W= m x g x h
h= W/m x g
h= 784/ 20 x 9.8
h=784/196
h=4m
Therefore, the height through which the body was lifted=4m
Questions Based on High Order Thinking Skills(HOTS)
Q.71 A boy tries to push a truck parked on the roadside. The truck does not move at all. Another boy pushes a bicycle. The bicycle moves through a certain distance. In which case was the work done more: on the truck or on the bicycle? Give a reason to support your answer.
Sol) The boy pushing the truck does no work as displacement is zero(no motion of the truck). But, when the boy pushes the bicycle, work done is more as the force applied by him causes displacement (motion of the bicycle).
Q.72 The work done by a force acting obliquely is given by the formula W = F cosθ x s. What will happen to the work.................... is increased gradually? Will it increase, decrease or remain constant?
Sol. On increasing the value of angle between force and distance, the value of cos θ decreases to zero when θ=90∘Therefore, work done decrease as angle increase, work done is 0 when θ=90∘ and work done is negative when angle further increases.
cos θ reduces further to –1.
Therefore, work done decrease as angle increase, work done is 0 when θ=90∘ and work done is negative when angle further increases.
Q.73 What should be the angle.................... of a body so that the work done is zero?
Sol) When the force is perpendicular to the direction of motion,work done will be zero. i.e., the angle should be 90∘ between force and displacement, for the work to be zero.
Q.74 In which of.................... when the angle between the direction of force and direction of motion is 0° or 90°?
Sol. When angle is 0∘
Q.75 How much work is done by the.................... moving around it in a circular path? Give reason for your answer.
Sol) Work done is zero, as force and displacement are at 90∘ to each other.
Q.76 A man is.................... of a hill at a height of 1200 meters. The man weighs 800 N and the package weighs 200 N. If g =10 m/S2.
(i) How much work does man do against gravity?
(ii) What is the potential energy of the package at A if it is assumed to be zero at B?
Sol. (i) Weight of the man= 800N
Weight of the package= 200N
Total weight= 800 + 200= 1000N
Height of the hill= 1200m
g=10m/s2
(ii)Given, Weight of the package(mg)= 200N, h=1200m
P.E = m x g x h
= 200× 1200
=2.40000J
=2.4 KJ
Q.77 When a ball is thrown vertically upwards,.................... What happens to its potential energy as its velocity becomes zero?
Sol) The potential energy keeps increasing as velocity keeps decreasing. It is maximum when velocity is zero.
Q.78 A man X goes to the top of a building by a vertical spiral staircase. Another man Y of the same mass.................... ladder. Which of the two does more work against gravity and why?
Sol. Work done against gravity depends on the vertical distance through which the body is lifted and is independent of the path travelled to achieve the height. Therefore, we can say that both men did the same work as the vertical distance travelled by them is the same.
Q.79 When a ball is thrown inside a moving bus, does its kinetic.................... the bus? Explain.
Sol. Yes. The kinetic energy of the ball thrown inside a moving bus is dependent on the speed of the moving bus. This is because the speed of the bus adds up to the speed with which the ball is thrown inside the moving bus.
For example, If the bus is moving at a speed of 30 km.hr and a person inside the bus throws a ball with a speed of 40 km/hr in the same direction as that of the motion of the bus. Then, for a person inside the bus, the K.E= ½ x m x (30)2. But, for a person who is static and is outside the bus, the K.E of the ball=1/2 x m x (30+ 40)2. Thus, the motion of the bus affects the K.E of the ball.
Q.80 A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy ? The bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to kinetic energy originally in the bullet?
Sol. Given, mass= 15g = 0.015kg; Initial velocity= 400m/s, distance, s=2cm=0.02m
K.E(initial)=1/2mv2
= ½ x 0.015 x (400)2
K.E(initial) = 1200J
As final velocity = 0.
Therefore, K.E(final)=0 ( Where F is the average force )
F=1200/ 2×10−2= -6×104N.
Negative sign shows that force is acting opposite to the direction of motion of the bullet. The kinetic energy originally in the bullet is eventually converted to heat energy by friction.
Very Short Answer Type Questions:
Q1) Name the.................... of energy.
Sol. KWh
Q2) Define kilowatt-hour.
Sol) 1 Kilowatt hour is the energy consumed when 1000 Watt is consumed for 60 minutes.
Q3) Name two units of.................... than watt
Sol) Kilowatt (KW) and Megawatt (MW)
Q4) Define the term 'watt'.
Sol. One Watt is defined as the power when one Joule of work is done in 1 s.
Q5) How many.................... one horse power?
Sol. 746 watt
Q6) Name the.................... whose unit is watt.
Sol. Power
Q7)What is the power of.................... the rate of one joule per second ?
Sol. 1 W
Q.8) A body does 1200 joules of work in 2 minutes. Calculate its power.
Sol. work =1200 J, time= 2 min= 2x 60 sec= 120 sec
Power = work done/time taken
= 1200/120
=10 W
Q9) How many joul.................... one kilowatt-hour ?
Sol. 3.6×106J.
Q10) Name the quantity whose unit is :
(a) kilowatt (b) kilowatt-hour
Sol. (a) Power (b) Energy.
Q.11) What is the common name of '1 kWh' of electrical energy ?
Sol. Unit of electricity
Q12) A cell converts.................... form. Name the two forms.
Sol. A cell converts Chemical energy into Electrical energy.
Q13) Name the device which.................... .
Sol. Electric motor
Q14) Name the devices or machines which convert :
(a) Mechanical energy into Electrical Energy.
(b) Chemical energy into Electrical Energy.
(c) Electrical energy into Heat Energy.
(d) Light energy into Electrical Energy.
(e) Electrical energy into Light Energy.
Sol)
(a) Electric generator (b) Battery
(c) Electric iron (d) Solar cell. (e) Electric bulb
Q15)Name the devices or machines which convert :
(i) Electrical energy into sound energy.
(ii) Heat energy into kinetic energy (or mechanical energy).
(iii) Chemical energy into kinetic energy (or mechanical energy).
(iv) Chemical energy into heat energy.
(v) Light energy into heat energy.
Sol) (i) Speaker in radio and television. (ii) Steam engin
(iii) Car engine. (iv) Gas stove (v) Solar water heater.
Q16) Fill in the following blanks with suitable words :
(a) Power is the rate of doing.......
(b) 1 watt is a rate of working of one....per.....
(c) The electricity meter installed in our home measures electric energy in the units of......
(d) The principle of......of energy says that energy can be......from one form to another, but it cannot be......or......
(e) When a ball is thrown upwards......energy is transformed into.......energy.
Sol. (a) Power is the rate of doing work
(b) 1 watt is a rate of working of one joule per second
(c) The electricity meter installed in our home measures electric energy in the units of kWh
(d) The principle of conservation of energy says that energy can be transformed from one form to another, but it cannot be created or
(e) When a ball is thrown up wards kinetic energy is transformed into potential energy.
Short Answer Type Questions:
Q17) A trolley is pushed a.................... in 1 minute. Calculate the power developed.
Sol. Force= 400N, distance travelled= 60m, t= 1 min=60 s
Work done= Force x distance
Work done= 400 x 60
Work done= 24000 J
Power = Work done/Time
Power= 24000/60
= 400 W.
Q18) What kind of energy.................... power station?
Sol. Potential energy of water → K.E of water → K.E of turbines → Electrical Energy.
Q19) What kind of.................... coal-based thermal power station?
Sol. Chemical energy → Heat energy → Kinetic energy of steam → K.E of turbines → Electrical Energy.
Q20) A man weighing.................... 4 m high in 5 seconds. What is the power?
Sol) Given, weight of the man= 500N
Weight of the load=100N
Total weight= 500 + 100= 600N
Height(vertical distance)= 4m
Work done= total weight x vertical distance
Work done= 600 x 4
Work done= 2400 J
Power = Work done/time
=2400/ 5=480W.
Q21) The power output of an engine is 3 KW. How much.................... in 20 s?
Sol) Given, Power= 3KW=3000W; t=20s
= 3000 × 20 = 60000 J = 60 KJ.
Work done =60KJ
Q.22) An electric.................... in 5 minutes. Calculate its power rating.
Sol. Energy= 600KJ= 600000J; time= 5 min= 5 x 60= 300s
Power = Energy/ Time=600000/300
Power= 2000W
Power of electric heater = 2000W
Q23)How much.................... does a 100 Watt lamp consume :
(a) in 1 second ? (b) in 1 minute ?
Sol) Given, power (P)= 100 watt
t=1 s
Energy = P × T = 100 × 1 = 100 J
(b) t=1min= 60 sec
Energy = 100 × 60 = 6000 J.
Q24) Five electric fans of 120 watts.................... the electrical energy consumed in kilowatt-hours.
Sol. Given, power= 120 watts=0.12 kw; t=4 hr
Energy consumed by one fan = 0.12 × 4 = 0.48 KWh.
Therefore, energy consumed by 5 fans= 0.48 x 5=2.4 kwh
Q25) Describe the energy changes which take place in a radio.
Sol) Electrical Energy →Kinetic Energy→ Sound Energy.
Radio converts electrical energy into kinetic energy and finally into sound energy
Q26) Write the energy.................... in an electric bulb (or electric lamp).
Sol) Electrical Energy → Heat Energy → Light Energy.
Q 27) Name five appliances or machines which use an electric motor.
Sol) Electrical fans, washing machine, refrigerator, grinder, electric iron.
Q 28)A bulb lights up.................... State the energy change which takes place:
(i) in the battery. (ii) in the bulb.
Sol)
(i) Chemical energy → Electrical energy.
(ii) Electrical Energy → Heat Energy → Light Energy
Q29)The hanging bob of a simple pendulum.................... released. It swings towards center position A and then to the other extreme position C.In which position do the bob have:
(i) maximum potential energy?
(ii) Maximum kinetic energy? Give reasons for your answer.
Sol. (i) Maximum P.E at the extreme position B and C because at these positions, the bob is at maximum height from the ground. K.E at these points is zero.
(ii) Maximum K.E at A because at these position, the bob has maximum velocity. Potential energy at this point is zero.
Q30) A car of weight 20000 N.................... 8 m/s, gaining a height of 120 m in 100 s. Calculate:
(a) Work done by the car.
(b) Power of engine of car.
Sol) (a) Weight of the car(W)= 20000 N
Height=120 m
Time taken (t)= 100 s
Work done by car = F × d =20,000 × 120 = 2400000 KJ or 2.4 x 106J
(b) Power = work done/ time=2400000/ 100=24 KW.
Long Answer Type Questions:
Q31) (a) What do you.................... term "transformation of energy”? Explain with an example.
(b) Explain the.................... following cases :
(i) A ball thrown upwards.
(ii) A stone dropped from the roof of a building.
Sol.
(i) K.E → P.E
(ii) P.E → K.E
Q.32
(a) State.................... of energy with an example.
(b) Explain how, the total energy.................... remains conserved. Illustrate your answer with the help of a labeled diagram.
Sol.
Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.Law of conservation of Energy states that the total energy of a system remains unchanged before and after transformation.
For example- a) A rock on a cliff. The rock has potential energy, when it's pushed off from a height, the potential energy is changed into kinetic energy.
The sum of potential energy and kinetic energy remains constant at every point of the falling of object.
mgh + ½ mv2= constant at every point
The sum of potential energy and kinetic energy is the total mechanical energy of the rock falling from a height(cliff).
During free fall of the object, the potential energy starts decreasing and converting into kinetic energy with decrease of height from the ground
(b) A swinging pendulum is a perfect example to show the conservation of energy. It shows the transformation of potential energy into kinetic energy and kinetic energy back into potential energy without any energy loss. In a pendulum, the law establishes that, when the ball is at its highest point, all the energy is potential energy and there is zero kinetic energy. At the ball's lowest point, all the energy in the ball is kinetic and there is zero potential energy. The total energy of the ball is the sum of the potential energy and kinetic energy.
Initially, the bob of the pendulum is at the mean position(B). When we draw the pendulum bob to one side(Extreme position A), we raise the bob to a little height end give it potential energy. This is the energy transferred by work done by hand. As at the extreme position, the bob has only PE, it tends to move down.The P.E decreases and K.E increases. At the lowest(mean) position, the bob has got K.E. Due to this it moves to the other side. Now, its K.E decreases and P.E increases. At the extreme positions A and C, all energy is in the form of potential energy and therefore it tends to move down. Thus the bob oscillates. At all other intermediate positions, energy of the pendulum is partly potential and partly kinetic. But, the total energy of the pendulum remains conserved.
Q33) (a) What is the meaning of the symbol KWh? What quantity does it represent?
(b) How much electric energy in KWh.................... of 1000 watts when it is switched on for 60 minutes?
Sol) (a) KWh means kilowalt hour. It is the commercial unit of measurement of electrical energy. 1 kWh = 3.6·106 J. It can be defined as the amount of electrical energy consumed when an electrical appliance of 1KW power runs for one hour.
(b) Power= 1000 watts(1KW); time= 60 min=1 hr
Energy = P × t = 1000 × 1h = 1000 Wh = 1 KWh.
Q34) (a) Derive the relation between commercial unit of energy (KWh) and SI unit of energy (joule).
(b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joules ?
Sol) (a) 1kWh = 1000 W × 1h.
1 watt= 1 joule/1 sec
= 1000 × 60 × 60
= 3.6×106J
Q34)(a) Derive the relation between.................... (KWh) and SI unit of energy (joule).
(b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joules ?
Sol)(a) 1kWh = 1000 W × 1h.
1 watt= 1 joule/1 sec
= 1000 × 60 × 6 = 3.6×106J
(b) Gn, total units consumed= 650 units
1 unit of electricity = 1 KWh(i.e., KWh is also known as unit)
650 units = 650 KWh. = 650 × 3.6 × 106J
= 2.34 × 109J
Q35)(a) Define power. Give the SI unit of power.
(b) A boy weighing 40 kg carries.................... 15 m high in 25 seconds. Calculate the power. (g = 10m/s2)
Sol) (a) Power is the rate of doing work . It is equivalent to an amount of energy consumed per unit time. In the SI system, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the steam engine.
1 Watt is the power of an object which does work at the rate of 1 Joule per second.
(b) Gn, mass of boy= 40Kg
Mass of box=20 kg
Total mass= 40+20=60kg
g = 10m/s2
h= 15m
t=25 sec
Power =mgh/t =60×10×15/ 25=360 W
Power= 360W
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