# Work, Energy and Power ICSE Class 10 Physics Important Questions

Students of ICSE Class 10 should refer to Work, Energy and Power ICSE Class 10 Physics Questions below which have come in past board exams. You should always go through questions that have come in previous years. This will help you to understand the pattern of questions in ICSE Class 10 Physics and prepare accordingly. This will help you to get better marks in ICSE Class 10 Board Exams

## ICSE Class 10 Physics Work Energy and Power Important Questions

Students should learn the important questions and answers given below for Chapter Work Energy and Power in Physics for ICSE Class 10. These board questions are expected to come in the upcoming exams. Students of ICSE Class 10th should go through the Important questions and answers ICSE Class 10 Physics which will help them to get more marks in exams.

### Work Energy and Power ICSE Class 10 Physics Questions

Work, Energy and Power ICSE Class 10 Physics Questions

SCOPE OF SYLLABUS

Definition of work, W = FScosq; special cases of q = 90º, 0º. W = mgh. Definition of energy, energy as work done. Various units of energy and work and their relation with S.I. Units. [ erg, calorie, kWh and eV]. Definition of Power, P = W/t; S.I. and C.G.S. units; other units, kilowatt (kW), megawatt (MW), and gigawatt (GW); and horse power (1H.P. = 746W), Simple numerical problems on work, energy and power. Mechanical Energy: U = mgh(Derivation included), Gravitational potential energy, examples; K = 1/2 mv2(Derivation included); forms of K.E.: Translational, Rotational and Vibrational – only simple examples. [Numerical problems on K and U only in case of translational motion]; qualitative discussions of electrical, chemical, heat, nuclear, light and sound energy, conversion from one form of energy to another; common examples. Statement of the principle of conservation of energy; theoretical verification that K + U = constant for a freely falling body. Application of this law to simple pendulum; simple numerical problems.

1. Define work. When does a force do work?
Ans. When force is applied on a body and there is displacement of the body, work is said to be done.

2. How is the work done by a force is measured when (i) force is in direction of displacement (ii) force is at an angle to the direction of displacement?
Ans. (i) Work = force × displacement in the direction of force
W = F × S
(ii) Work done = force × displacement of body
W = F × S cosθ

3. A force F acts on a body and displaces it by a distance S in a direction an angle θwith the direction of force (a) Write the expression for the work done by the force (b) What should be the angle between the force and displacement to get (i) zero (ii) maximum work?
Ans. (a) Work done = Force × Component of Displacement of body in the direction of force.
∴ W = F S cosθ
(b)
(i) 900 (ii) 00

4. A body is acted upon by a force. State two conditions when the work done is zero.
Ans. (i) There should be no displacement i.e. S = 0
(ii) The displacement is normal to the direction of force i.e θ = 900

5. State the condition when the work done by a force is (i) Positive (ii) Negative. Explain with the help of examples.
Ans. (i) Positive work is when displacement is in the direction of Force.
i.e. θ=00, cos 00 =1
∴ W = F × S
Example: If the car is displaced in the direction in which the force is applied.
(ii) Negative work is when the displacement is in a direction opposite to the Force.
i.e θ=1800, cos 1800 =-1
∴ W = -F × S
Example: When a ball is thrown upwards the work done by force of gravity is Negative.

6. A body is moved in a direction opposite to the direction of force acting on it.
State whether the work is done by the force or work is done againt the force.
Ans. Work is done against the force.

7. When a body moves in a circular path, how much work is done be the body?
Give reason.
Ans. No work is done, since displacement is normal to the direction of force on the body.

8. A satellite revolves around the earth in a circular orbit. What is the work done by the satellite? Give reason.
Ans. The work done by the satellite is zero. This is because the force of gravity is directed towards the centre of circular path of the satellite i.e. Earth and the displacement at all instants is along the tangent to the circular path i.e. normal to the direction of fore on satellite.

W = F × S cos 900
W = F × S × 0 = 0

O – Earth
S – Displacement
F – Force of gravity

9. In which of the following cases, is work being done?
(i) A man pushing a wall
(ii) A coolie standing with a load of 12 kg on his head
(iii) A boy climbing up a staircase
Ans. (iii) A boy climbing up a staircase.

10. A coolie X carrying a load on his head climbs up a slope and another coolie Y carrying the identical load on his head moves the same distance on a frictionless horizontal platform. Who does more work? Explain the reason.
Ans. Coolie X carrying a load on his head work against force of gravity while coolie Y moves on a frictionless horizontal platform, Y does not work because displacement is normal to the force of gravity.

11. The work done by a fielder when he takes a catch in a cricket match, is negative. Explain.
Ans. The work done by a fielder while catching a ball is negative because he applies force against the direction of ball.

12. Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaced from its initial position.
Ans. A person carrying a box on his head moving on ground does no work against the force of gravity as displacement of load (box) being horizontal, is normal to the direction of force of gravity.

13. What are the S.I and C.G.S units of work? How are they related? Establish the relationship.
Ans. S.I. unit of work is joule (J). C.G.S unit of work is erg.
Relationship:
1 joule = 1N × 1m
1 joule = 105 dyne × 100 cm = 107 dyne × cm
∴ 1 joule = 107 erg

14. State and define the S.I. unit of work
Ans. S.I unit of work is joule.
Joule is that much work done when a force of 1N displaces the body through a distance of 1m in the direction of force.

15. Express joule in terms of erg.
Ans. 1 joule = 107 erg

16. A body of mass m falls down through a height h. Obtain an expression for the work done by the force of gravity.
Ans. Let a body A of mass ‘m’ be at a height ‘h’ above the ground level B falls down. Force acting on at A is mg force of gravity.
Displacement = height = h
Work done in falling W = force × displacement
W = mg × h
Work done (W) = mgh.

17. A boy of mass m climbs up a staircase of vertical height h.
(a) What is the work done by the boy against the force of gravity?
(b) What would have been the work done if he uses a lift in climbing the same vertical height?
Ans.     (a) W = mgh       (b) W = mgh

17. A boy of mass m climbs up a staircase of vertical height h.
(a) What is the work done by the boy against the force of gravity?
(b) What would have been the work done if he uses a lift in climbing the same vertical height?
Ans. (a) W = mgh (b) W = mgh

18. Define the term energy and state its S.I unit.
Ans. Capacity of doing work is called Energy. S.I. unit: joule (J).

19. What physical quantity does the electron volt (eV) measure? How is it related to the S.I unit of that quantity?
Ans. The physical quantity which electron volt measures is Energy.
Relation: 1 eV = 1.6 × 10-19 J

20. Complete the following sentence:
(a) 1 J = ……. calorie. (b) 1 kWhr = ………. J
Ans. (a) 0.24 (b) 3.6 × 106

21.Name the physical quantity which is measured in calorie. How is it related to the S.I unit of that quantity?
Ans. Physical quantity is Heat energy.
S.I Unit of Heat Energy: joule
joule is related to Calorie as 1 J = 0.24 calorie or 1 calorie = 4.18 J

22. Define a kilowatt hour. How is it related to joule?
Ans. 1 kWh is the work done by a source of power 1 kW in 1 h.
1 kilowatt hour = 3.6 ×106 J

23. Define the term ‘power’. State its S.I unit.
Ans. The rate of doing work is called power.
S.I. unit of power is watt.

24. State two factors on which power spent by a source depends. Explain your answer with examples.
s. (i) The amount of work done by the source.
(ii) The time taken by the source to do the said work.

Example :- If a coolie A takes one minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of same bus, then work done by both the coolies is the same, but the power spent by the coolie B is twice because the coolie A does work at the faster rate

Differentiate between work and power:

26. Differentiate between energy and power.
Ans.

27. State and define the S.I unit of power
Ans. S.I unit of power is Watt.
Power is said to be one watt of 1 joule of work is done in 1 sec.

28. What is horse power (H.P)? How is it related to the S.I unit of power?
Ans. Horse power [ H.P] is the power of an agent which can work 746 joules per second.
Relation: 1 H.P. = 746 W = 0.746kW

29. Differentiate between watt and watt hour.
Ans. If 1 joule of work is done in 1 second, the power spent is said to be 1 watt. It is S.I.
unit of power. Whereas watt hour is bigger unit of energy it is the power of 1 watt
which is spent in 1 hour = 3.6kJ.

30. Name the quantity which is measured in
(a) kWh    (b) kW (c)Wh (d) eV
Ans.
(a) kWh – Energy
(b) kW – Power
(c) Wh – Energy
(d) eV – Energy

MULTIPLE CHOICE QUESTIONS

1. One joule work is said to be done when:
(a) A force of 1 N displaces a body by 1 cm
(b) A force of 1 N displaces a body by 1 m
(c) A force of 1 dyne displaces a body by 1 cm
(d) A force of 1 dyne displaces a body by 1 m

(b) A force of 1 N displaces a body by 1 m

2. kilowatt is the unit of:
(a) work
(b) momentum
(c) force
(d) power

Ans. (d) power

3. One horse power is equal to
(a) 1000 W
(b) 500 W
(c) 764 W
(d) 746 W

(d) 746 W

4. kWh is the unit of :
(a) power
(b) force
(c) energy
(d) none of these

(c) energy

NUMERICALS BASED ON WORK, ENERGY AND POWER

1. A body, when acted upon by a force of 10kgf, gets displaced by 0.5m. Calculate the work done by the force, when the displacement is (i) in the direction of force (ii) at an angle of 600 with the force (iii) normal to the force. (g = 10 N kg-1)

Ans. F = 10kgf = 10 × 10N = 100 N, S = 0.5m
(i) Work done when displacement is in the direction of force
W = F × S = 100 × 0.5 = 50J
(ii) Work done when displacement is at the angle of 600
W = F.S cos 600 = 100 × 0.5 × 1/2 = 25J
(iii) Work done when displacement is normal to the direction of force
W = F.S cos 900 = 100 × 0.5 × 0 = 0 J

2. A boy of mass 40 kg runs upstairs and reaches the roof at height 8m high in 5 s. Calculate:
(i) The force of gravity acting on the boy
(ii) The work done by him against gravity
(iii) The power spent by boy. (Take g = 10 ms-2)

Ans. m = 40 kg, h = 8m, t = 5s
(i) F = mg = 40 × 10 = 400 N
(ii) W = F × h = 400 × 8 = 3200 J
(iii) P = m ´ g ´ h /t = 40 ´ 10 ´ 8/5 = 640 W

3. It takes 20s for a person A to climb up the stairs, while another person B does the same in 15s. Compare the (i) work done and (ii) power developed by the person A and B

Ans. As height in both cases is same
(i) Work done by A = Force × height = mgh
Work done by B = Force × height = mgh
1 : 1 (provided masses of both A and B are same)

4. A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute. Calculate (i) the work done (ii) power spent.
Ans. Force (F) = weight of body = 350 N
Total distance (h) = 30 × 20 = 600 cm = 600/100 = 6m
(i) Work done (W) = F × h = 350 × 6 = 2100J

(ii) Power spent = Work done /Time taken = 2100/60 = 35 W

5. A man spends 6.4 kJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5s. Calculate: (i) the force applied (ii) the power spent (in H.P) by the man.

Ans. Energy = Work done = 6.4 kJ = 6.4 × 1000 = 6400J, S = 64m, t = 2.5s
(i) F = ?
Work done = Force × distance
6400 = F × 64
F = 6400 /64 = 100 N
(ii) P = W /t = 6400/2.5 = 2560 watt
P (H.P.) = 2560 /746 = 3.43 H.P.

6. A weight lifter lifted a load of 200 kgf to a height of 2.5 m in 5 s. Calculate:

(i) the work done (ii) the power developed by him. (Take g = 10 N kg-1)

Ans. Load = force = 200 kgf = 200 × 10 = 2000 N, h = 2.5m, t = 5s, g = 10N kg-1
(i) Work done = F × h = 2000 × 2.5 = 5000 J
(ii) Power = W /t = 5000 /5 = 1000 W

7. A machine raises a load of 750 N through a height of 16 m in 5 s. Calculate:
(i) energy spent by machine (ii) power at which the machine works
Ans. Load = Force = mg = 750 N, h = 16 m, t = 5s
(i) Since, energy spent is its capacity to do work
Work done = mgh = 750 × 16 = 12000 J
(ii) Power of machine = mgh /t = 12000 /5 = 2400 W

8. An electric heater of power 3 kW is used for 10 h. How much energy does it consume? Express your answer in (i) kWh (ii) joule
Ans. Power of heater (P) = 3 kW, Time (t) = 10 h
(i) Energy consumed by heater = power × time = 3 × 10 = 30kWh
(ii) 1 kWh = 3.6 × 106J
30kWh = 30 × 36/ 10 × 106J = 108 ×106J

9. A boy of mass 40 kgf runs up a flight of 15 steps, each 15 cm high in 10s.
Find the work done and the power developed by him. Take g = 10 Nkg-1
Ans. m = 40 kg, h = 15 × 15 = 225 /100 = 9/4 m, g = 10 Nkg–1
work done = mgh = 40 × 10 × 9 /4 = 900J
Power developed = mgh / t = 900 /10 = 90W

10. A water pump raises 50 litres of water through a height of 25 m in 5s.
Calculate the power which the pump supplies.
(Take g = 10 N kg-1 and density of water = 1000 kg m-3 )

Mass of 50 litres of water = 50kg

h = 25 m, t = 5 s, g = 10Nkg-1
Power supplied by pump =
Mgh / t = 50 ×10 × 25 /5 = 2500W

11. A man raises a box of 50 kg mass to a height 2 m in 2 minutes, while another man raises the same box to the same height in in 5 minutes. Compare:
(i) the work done and (ii) the power developed by them.
Ans. m = 50kg, h = 2m
(i) Work done by each man = mgh = 50 × 10 × 2 = 1000J
Work done is independent of time taken
Ratio of work done = 1 : 1

12. A pump is used to lift 500 kg of water from a depth of 80 m in 10s.
Calculate:
(a) The work done by the pump
(b) The power at which the pump works
(c) The power rating of the pump if its efficiency is 40%.
(Take g = 10 ms-2)

Ans. m = 500 kg, h = 80m, t = 10s
(a) Work done by pump = mgh = 500 × 80 × 10 = 400000 J
(b) Power = Work done / Time = 400000 /10 = 40000W
(c) Efficiency of pump = 40%
Efficiency = useful power /power input
40 = 40000 /power input × 100
power input = 40000´ 100 /40 = 100000 W

13. 1 An ox can apply a maximum force of 1000 N. It is taking part in a cartrace and is able to pull the cart at a constant speed of 30 ms-1 while making its best effort. Calculate the power developed by the ox.

Ans. F = 1000N, v = 30 ms-1
P = F × v =1000 × 30 = 30kW

If the power of a motor is 40 kW, at what speed can it raise a load of 20,000 N ?
Ans.
power = Force × speed
Speed = power /Force = 40000/20000 = 2ms-1

DIFFRENT FORMS OF ENERGY

1. What are the two forms of mechanical energy?
Ans. (i) Kinetic Energy (K) (ii) Potential Energy (U)

2. Name the form of energy which a wound up watch spring possesses.
Ans. Elastic Potential Energy

3. Name the type of energy (kinetic energy K or potential energy U) possessed in the following cases:
(a) A moving cricket ball
(b) A compressed spring
(c) A moving bus
(d) The bob of a simple pendulum at its extreme position.
(e) The bob of a simple pendulum at its mean position.
(f) A piece of stone placed at the roof.

Ans. (a) K (b) U (c) K (d) U (e) K (f) U

4. When an arrow is shot from a bow, it has kinetic energy in it, explain briefly from where does it get its kinetic energy?
Ans. Arrow has mass and velocity is produced in an arrow when it is shot. It gains kinetic energy due to motion.

5. Define the term potential energy of a body. State its different forms and give one example of each.

Ans. The energy possessed by a body because of its Position or its state is called its Potential Energy.
Two forms of Potential energy are
(i) Elastic potential energy. Ex.: A compressed spring
(ii) Gravitational potential energy. Ex.: Water stored in dam

6. A ball is placed on a compressed spring. What form of energy does the spring possess? On releasing the spring, the ball flies away. Give a reason.

Ans. A compressed spring possess (P.E) Potential energy due to changed shape. When it is released, the P.E of spring, changes into (K.E) Kinetic energy which does work on the ball placed on it and makes the ball fly away.

7. What is meant by the gravitational potential energy? Derive expression for it.

Ans. Gravitational potential Energy is the potential energy possessed by a body due to of the earth on it.
Derivation of Expression :
Let a body of mass m be lifted from ground level B
to A through height h. For this work has to be done.
W = Force of gravity ´ displacement = m x g x h
This work done is stored in the body as
Gravitational potential
Energy [U = mgh] .

8. Write an expression for the potential energy of a body of mass m placed at a height h above the earth’s surface.
Ans. Potential Energy = m × g × h

9. Name the form of energy which a body may possess even when it is not in motion. Give an example to support your answer.
Ans. Chemical Energy, Heat Energy, Magnetic Energy, Eletrical Energy.

10. What do you understand by the kinetic energy of a body?

Ans. The energy possessed by a body, by virtue of its motion is called Kinetic Energy.

K.E.= 1/2 mv2

11. A body of mass m is moving with a velocity v. Write the expression for its kinetic energy.
Ans. Kinetic energy = 1 /2 m x v2

12. State the work-energy theorem.

Ans. When force is applied on the moving body in the direction of force (a) work is done (b) Increase in K.E takes place which is equal to the work done by the force. This is known as Work Energy Theorem.

13. A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v. How much work is being done by the force?

Ans. When the body is at A
Work done by the force = Force x displacement
W = F ´ S …(i)
From relation v2 = u2 + 2aS,
S= v2 -u2 /2a
and F = ma

From equation (i),
W = ma ´ v2 -u2 /2a
W = ½ m(v – u ) ……(ii)
Initial K = K1 = 1/2mu2
Final K = K2 = 1/2mv2

Increase in K.E. = K2 – K1 = ½ mv2 – 1/2 mu2 = 1/2 m  (v2– u2)………(iii)
from equation (ii) and (iii)
work done = increase in K.E
W = 1/2 mv2 – 1/2 mu2

14. A light mass and a heavy mass have equal momentum. Which will have more kinetic energy?

Ans. momentum = mass x velocity
Let m1,m2 be the mass of two bodies having velocity v1,v2
∴ Their momentum is p1 = m1.v1 and p2 = m2.v2 respectively.
Since, m1v1 = m2v2
If m1 < m2 , then v1 > v2
Kinetic Energy of lighter mass is more.

15. Name the three forms of kinetic energy and give one example of each.
Ans. (1) TRANSLATIONAL KINETIC ENERGY
Examples:
(a) A freely falling body.
(b) A car moving in a straight line
.
(2) VIBRATIONAL KINETIC ENERGY

Examples :
(a) In a solid, the atoms vibrate about their mean position and so they possess the Vibrational Kinetic Energy
(b) A steel blade clamped at one end when displaced slightly at the other end and then released starts vibrating and is said to possess the Vibrational Kinetic Energy.

(3) ROTATIONAL KINETIC ENRGY
Examples :
(a) Rotating earth
(b) A rotating wheel

16. Differentiate between potential energy (U)and kinetic energy (K).
Ans.

17. Complete the following sentences:
(a) The kinetic energy of a body is the energy by virtue of its……..
(b) The potential energy of a body is the energy by virtue of its……
Ans. (a) motion (b) state or position

18.Is it possible that no transfer of energy may take place even when a force is applied to a body?
Ans. Yes, when force is normal to displacement.

19.Name the form of mechanical energy, which is put to use.
Ans. Kinetic energy

20. In what way does the temperature of water at the bottom of a waterfall differ from the temperature at the top? Explain the reason.
Ans. When water falls from a height, the potential energy stored in water at the height changes into kinetic energy of water during the fall. On reaching the bottom the kinetic energy of water changes into heat energy due to which the temperature of water at the bottom rises.

DIFFRENT FORMS OF ENERGY & ITS CONVERSION

1. What are the different forms of energy ? Name them.
Ans. (i) Solar Energy
(ii) Chemical Energy
(iii) Electrical Energy
(iv) Heat Energy
(v) Light Energy
(vi) Sound Energy
(vii) Hydro-Energy
(viii) Geothernal Eeergy
(ix) Wind Energy
(x) Mechanical Energy
(xi) Nuclear Energy
(xii) Magnetic Energy

2. Energy can exist in several forms and change from one from to another. For each of the following, state the energy changes that occur in:
(a) The unwinding of a watch spring
(b) a loaded truck when started and set in motion
(c) a car going uphill
(d) photosynthesis in green leaves
(e) charging of a battery
(f) respiration
(g) burning of a match stick
(h) explosion of crackers
Ans. (a) Potential Energy to Kinetic Energy
(b) (i) Electrical energy to Chemical energy
(ii) Chemical energy to Mechanical energy
(c) (i) Chemical energy to Kinetic energy
(ii) Chemical energy to Potential energy
(d) Light energy to Chemical energy
(e) Electrical energy to chemical energy
(f) Chemical energy to heat energy
(g) Chemical energy to heat energy
(h) Chemical energy to sound energy

3. State the energy changes in the following cases while in use:
(a) loudspeaker
(b) microphone
(c) a glowing electric bulb
(d) a solar cell
(e) an electric cell in a circuit
(g) a petrol engine of a running car.
(h) an electric toaster
(i) a photovoltaic cell
(j) an electromagnet
Ans. (a) Electrical energy to Sound energy
(b) Heat energy to Rotational energy
(c) Sound energy to Electrical energy
(d) Electrical energy to Rotational energy
(e) Electrical energy to Light and Heat energy
(f) Chemical energy to Heat energy
(g) Light energy to Electrical energy
(h) Chemical energy to Heat energy
(i) Chemical energy to Electrical energy
(j) Chemical energy or Heat energy to Mecanical energy
(k) Electrical energy changes to Heat energy
(l) Photovoltaic cell, Light energy to Electrical energy
(m) Electric energy changes to the Magnetic energy

MULTIPLE CHOICE QUESTIONS

1. A body at a height possesses:
(a) kinetic energy
(b) potential energy
(c) solar energy
(d) heat energy
Ans. (b) potential energy

2. The momentum p and kinetic energy K of a moving body are related as:
(a) p = 2m K
(b) p = K/2m
(c) K = p2/2m
(d)K = 2m p2
Ans. (c) K = p2/2m

3. When a body falls down, its potential energy from U1 to U2.They are related as:
(a) U1 = U2
(b) U1 < U2
(c) U2 < U1
(d) noting can be said
Ans. (c) U2 < U1

4. In an electric cell while in use, the change in energy is from:
(a) electrical to mechanical
(b) electrical to chemical
(c) chemical to mechanical
(d) chemical to electrical
Ans. (d) chemical to electrical

NUMERICALS BASED ON MECHANICAL ENERGY

1. Two bodies of equal masses are placed at heights h and 2h.
Find the ratio of their gravitational potential energies.
Ans.
Gravitational Potential energyof A/ GravtiationalPotential energy of B
= mgh/ mg2h = 1 /2

2. Find the gravitational potential energy of 1 kg mass kept at a height of 5 m above the ground if g = 10ms-2

Ans. m = 1kg, h = 5m, P.E. = ?
Gravitational Potential energy = mgh = 1 x 10 x 5 = 50J

3. A box of weight 150 kgf has gravitational potential energy stored in it equal to 14700 J. Taking g = 9.8 Nkg-1, find the height of the box above the ground.
Ans. m = 150 kgf, P.E. = 14700 J, g = 9.8 Nkg-1, h = ?
P.E. = mgh
14700 = 150 x 9.8 x h
h = 14700 /150 × 9.8 = 10m

3.A body of mass 5 kg falls from a height of 10 m to 4 m. Calculate:
(i) The loss in potential energy of the body
(ii) The total energy possessed by the body at any instant?(Take g = 10 m s
-2)
Ans.
(i) Loss in Potential energy = P.E at height 10 m – P.E at height 4 m
= mgh1 – mgh2
= mg ( h1– h2 )
= 5 x 10 x (10 – 4) = 5 x 10 x 6 = 300J
(ii) Total energy of body at any instant = mgh = 5 ´ 10 ´ 10 = 500 J

5. Calculate the height through which a body of mass 0.5 kg should be lifted it the energy spent in doing so is 1.0 J. Take g = 10 ms-2.
Ans. P.E. = 1 J, m = 0.5 kg, h = ?, g = 10 ms-2
P.E.= mgh
1 = 0.5 x 10 x h
h = 1 /5 = 0.2m

6. A boy weighing 25 kgf climbs up from the first floor at height 3 m above the ground to the third floor at height 9 m above the ground. What will be the increase in his gravitational potential energy? (Take g = 10 N kg-1)

Ans. Force acting on body = 25 kgf = 25 x 10 = 250N
As height changes from 3m to 9m
Increase in P.E. in rising from 3m to 9m above the ground = mgh2 – mgh1
= mg [h2 – h1]
= 250 [ 9 – 3]
= 250 x 6 = 1500 J

7. A vessel containing 50 kg of water is placed at a height 15m above the ground.
Assuming the gravitational potential energy at ground to be zero, what will be the gravitational potential energy of water in the vessel ? (g = 10ms-2)

Ans. m= 50 kg, h = 15m, g = 10ms-2, P.E. = ?
∴ Gravitational Potential energy = m x g x h = 500 x 15J = 7500 J

8. A man of mass 50kg climbs up a ladder of height 10 m. Calculate:
(i) the work done by the man (ii) the increase in his potential energy. (g= 9.8 ms-2)
Ans.
m= 50 kg, g = 9.8 ms-2, h = 10m
(i) Work done by man = mgh = 50 x 9.8 x 10 = 4900J
Taking potential energy at ground = 0
= 50 x 9.8 [ 10 – 0]
= 50 x 9.8 x 10
= 4900J

9. A block A, whose weight is 200 N, is pulled up a slope of length 5 m by means of a constant force F(= 150 N) as illustrated in figure.
(a) What is the work done by the force F in moving the block A, 5 m along the slope?
(b) By how much has the potential energy of the block A increased?
(c) Account for the difference in work done by the force and the increase in potential energy of block.

Ans. (a) F = 150 N, S = 5m
Work done in moving the block along slope W = F x s
W = 150 x 5 = 750 J
(b) When block reaches the top, the work done is stored in it as P.E.
vertical height to top = 3m
P.E. at top = mgh = 200 x 3 = 600J
Increase in P.E. of body = 750 – 600 = 150J
(c) This difference of 150 J energy is used against friction between the block and slope which will appear as heat.

10. Find the kinetic energy of a body of mass 1 kg moving with a uniform velocity of 10ms-1.
Ans. m = 1 kg, v = 10ms-1
K.E. =1/2 mv2 =1/2 x 1 x 10 x 10 = 50 J

11. If the speed of a car is halved, by which factor does its kinetic energy change?

Ans. K.E. =1/2 mv2
When    v1 =1/2 v
∴   K.E. =1/2 mv21  =1/2 m

new  K.E. =1/4 K.E.
∴ K.E. changes to1/4 th of the original

12. Calculate the decrease in Kinetic energy of a moving body if its velocity is reduced to 1/3rd of the initial velocity.

Ans. Initial velocity = v; Final velocity 1/3 v= v/3
Initial kinetic energy K1 =1/2 x m x v2 =1/2 mv2
Final kinetic energy K2 =1/2 x m x v/3) 2 =1/2 ´1/9 mv2
Change in kinetic energy =1/2 x mv2 –1/2 ´1/9 mv2

13. Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.
Ans.
A                                                          B

K.E. = 1/2 mv2                                             K.E. = 1/2 m (2v)2  = (1/2mv2)x4
∴ Ratio of their K.E = 1/2mv2 / 4(1/2mv2) = ¼ =1:4

14. A car is running at a speed of 15 km h-1 while another similar car is moving at a speed of 30 km h-1. Find the ratio of their kinetic energies:

Ans. Two similar cars means having same masses
vA = 15km h-1, vB = 30 km h-1
Ratio of K.E.= ?
∴ Kinetic Energy of Car A/Kinetic Energy of Car B = { 1 / 2 m (15)2 } / { 1 / 2 m (30)2 } =15×15 / 30×30 = 1/4

15. A ball of mass 0.5 k.g slows down from a speed of 5 ms-1 to that of 3ms-1. Calculate the change in kinetic energy of the ball.

Ans. m = 0.5 kg, u = 5ms-1, v = 3 ms-1
Initial K.E. =1/2 mu2 =1/2 x 0.5 x 5 x 5 =25/4 J
Final K.E. =1/2 mv2 =1/2 x 0.5 x 3 x 3 =9/4 J
Change in K.E.=25/4 –9/4 =16/4 = 4J
∴ K.E. decreases by 4J

16. A cannon ball of mass 500 g is fired with a speed of 15 ms-1. Find: its kinetic energy and momentum.
Ans.
m = 500 g =500/1000 =1/2 kg, v = 15 ms-1
Kinetic energy (K.E) =1/2 mv2 =1/2 x 1/2 x 15 x 15 = 56.25J
Momentum energy of ball (p) = mv =1/2 x 15 = 7.5 kg ms-1

17. A bullet of mass 50g is moving with a velocity of 500 ms-1. It penetrates 10 cm into a still target and comes to rest. Calculate: (i) the kinetic energy possessed by the bullet (ii) the average retarding force offered by the target.

Ans. m = 50g =500/1000 =1/20 kg , v = 500ms-1
K.E. of bullet =1/2 mv2 =1/2 x1/20 x 500 ´ 500 = 6250 J
(b) u = 500 ms-1,v = 0 (as bullet comes to rest), S = 10 cm =101/00 =1/10 m
v2 – u2 = 2aS
0 – 500 x 500 = 2a ´110
∴                 a = –1250000 ms-2
∴  Retardation = 1250000 ms-2
But F = m x a
∴ Retarding force offered by target =1/20×1250000 = 62500 N
Average Retarding Force offered is 62500N.

18. A body of mass 10 kg is moving with a velocity 20ms-1. If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy.

Ans. m = 10kg, v = 20 ms-1
(i) K.E. = 1/2mv2 =1/2 x 10 x (20)2 = 2000J
If mass is doubled i.e. 20kg, velocity is halved i.e. 10ms-1
(i) K.E. = 1/2mv2 =1/2 x 20 x (10)2 = 1000J
∴ Ratio = 2000 : 1000 = 2 : 1

19. A truck weighing 1000 kgf changes its speed from 36 km h-1 to 72 km h-1 in 2 minutes. Calculate: (i) the work done by the engine and (ii) its power. (g = 10ms-2)
Ans. Weight of truck = F = 1000 kgf, m = 1000kg, u = 36 km h-1 = 36 x 5/18 = 10 ms-1 , v
= 72 km h-1 = 72 x 5/18 = 20 ms-1 , t = 2 minutes = 2 x 60 = 120s
(i)Work done = Increase in energy
= 1/2 mv2 –1/2 mu2 =1/2 m [(v + u) (v – u)]
=1/2 x 1000 (20 + 10) (20 – 10) =1/2 x 1000 x 30 x 10
= 150000 J
(ii) ∴ Power of Engine = W/t =150000 /120 = 1250W

20. A body of mass 60 kg has the momentum 3000 kg ms-1. Calculate: (i) kinetic energy (ii) speed of the body.

Ans. (ii) m = 60kg
p = 3000 kg ms-1
p = m x v
∴   60 v = 3000
∴       v =3000/60 = 50ms-1
(i) Kinetic Energy K.E. = 1/2 mv2 = 1/2 x 60 x 50 x 50 = 75000J

21. How much work is needed to be done on a ball of mass 50g to give it a momentum of 500 g cm s-1?

Ans. p = 500g cm s-1 , m = 50 g = 500/1000 = 1/20 kg
p = m x v
∴ 500 v = 50
∴ v =500/50 = 10cm s-1 =10/100 = 1/10 ms-1
work done (W) =1/2 x m x v2 =1/2 x 1/20 x 1/10 x 1/10 = 2.5 x 10-4J

22. How much energy is gained by a box of mass 20 kg when:
(a) A man carrying the box waits for 5 minutes for a bus?
(b) He runs carrying the box with a speed of 3ms-1 to catch the bus?
(c) He raise it by 0.5 m in order to place it inside the bus? (g = 10ms-2)

23. A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown in the figure. When the trolley is released it is found to move at a speed of 2 ms-1. What potential energy did the spring possess when compressed?

Ans. Mass of trolley = 0.5 kg, g = 10ms-2, v = 2 ms-1
v2 – u2 = 2gS
(2)2 – 0 = 2 x 10 x S
S =4/20 = 1/5 = 0.2m
P.E. = mgS = 0.5 x 10 x 0.2 = 1J