Download Question Bank Work, Power and Energy

Question Bank
Work, Power and Energy
1. Define work.
Ans. Work is said to be done, when a force or its component causes displacement
in the direction of force.
2. State the conditions for doing work.
Ans. (i) There must be a force or component of force acting on a body.
(ii) The force or its component must produce displacement in its own direction.
3. State the mathematical expression for work when displacement is produced in
its own direction.
Ans. W = F × S, where, W is the work done, F is the force applied and ‘S’ is the
displacement in the direction of force.
4. State the mathematical expression for work when the force acts at an angle to
the direction of displacement of body.
Ans. W = F cos θ × S, where W is the work done, F cos θ is the component of force
acting in the direction of displacement and ‘S’ is the displacement of the body.
5. How is work done measured?
Ans. Work done is measured as a product of force and the displacement caused by it
in its own direction.
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6. State and define unit of work in CGS system.
Ans. Unit of work in C.G.S. system is erg.
When a force of 1 dyne causes a displacement of 1cm, in its own direction,
the work done is said to be one erg.
7. State and define unit of work in SI system.
Ans. Unit of work in SI system is Joule.
When a force of 1 N causes a displacement of 1 m, in its own direction, the
work done is said to be 1 joule.
8. Derive a relation between joule and erg.
Ans.
1J =1N×1m
=
1 kgm
×m
s2
= 1000 g ×
100 cm × 100 cm
s2
gcm 2
= 10
s2
= 107 ergs.
7
9. What should be the angle between the force and the displacement to get
(i) minimum work, (ii) maximum work.
Ans. (i) For minimum work, the angle should be 90°.
(ii) For maximum work, the angle should be zero.
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10. A body is acted upon by a force, state two conditions when work done is zero.
Ans. (1) The force must not cause displacement in its own direction.
(2) The force must act at right angle to direction of displacement.
11. 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 done, when the angle between the direction of force
applied and the displacement of body is zero.
For example, when a horse pulls a cart or an engine pulls a train, the
positive work is done.
(ii) Negative work is done, when the angle between the direction of the force
applied and the displacement is 180° i.e., the displacement is in a direction
opposite to the force.
For example, when brakes are applied to a moving vehicle, the work done
by the applied force is negative. Similarly, when a ball is projected
vertically upward, the work done by the force of gravity is negative.
12. Select the correct choice :
The work done by a force on a body will be positive if the :
(a) body does not move
(b) body moves perpendicular to the direction of applied force.
(c) body moves along the direction of applied force.
(d) body moves opposite to the the direction of the applied force.
Ans. (c) body moves along the direction of the applied force.
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13. When a body moves in a circular path, how much work is done by the body?
Give a reason.
Ans. No work is done. It is because, the centripetal force acting on the body is
always at right angles to the displacement of the body along the circular path.
Thus, cos 90° = 0
So, in the expression
W = F cos 90° × S = 0 × S = Zero.
14. A satellite revolves around the earth in a circular orbit. What is the work
done by the force of gravity?
Ans. Work done by the force due to gravity is zero, because it acts at right angles
to the direction of displacement of the satellite.
15. In which of the following cases, is the work being done and why?
(i) A man pushing a wall.
(ii) A coolie standing with a load of 12 kgf on his head
(iii) A boy climbing up a staircase
(iv) A boy carrying a box on his head and sliding over a frictionless
horizontal surface.
Ans. In case (iii) work is done, because displacement is produced in the direction of
applied force. In cases (i), (ii) and (iv), no work is done as no displacement
occurs in the direction of applied force.
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16. Give an example when the 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 salesman carrying a suitcase in hand and walking on a horizontal road, does
no work against the force of gravity. It is because no displacement takes place
in the direction of force of gravity.
17. Is work a scalar or vector quantity?
Ans. Work is a scalar quantity.
18. One joule of work is said to be done when :
(a) a force of 1 dyne causes a displacement of 1 cm.
(b) a force of 1 dyne causes a displacement of 1 m.
(c) a force of 1 N causes a displacement of 1 cm.
(d) a force of 1 N causes a displacement of 1 m.
Ans. (d) a force of 1 N causes a displacement of 1 m.
19. A boy of mass m climbs up a staircase of vertical height h. (i) What is the
force acting on the boy? (ii) What is the work done by the boy against the
force of gravity? What would have been the work done, if he uses a lift in
climbing the same height?
Ans.
(i) Force acting on the boy = mg.
(ii) Work done by the boy against the force of gravity = mgh.
(iii) Work done, if lift is used = mgh.
20. Define the term “Power”.
Ans. The rate of doing work is called power.
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21. Write two mathematical expressions for power.
Ans. (i) Power =
Work
w
i.e., P = .
Time
t
(ii) Power = Force × Velocity i.e., P = F × v
22. State and define SI unit of power.
Ans. SI unit of power is watt.
When a work of 1 Joule is done in 1 second, the power is said to be one watt.
23. Express SI unit of power in terms of MLT.
1J
m2 1
= 1 kg 2 ×
1W=
s
s
s
Ans.
m 2 ML2
= 1 kg 3 = 3 or ML2 T –3
s
T
24. Name two bigger units of power and express them in watts.
Ans. (i) Kilowatt (kW) = 103 W
(ii) Megawatt (MW) = 106 W.
25. What do you understand by the term horse power? Express this unit in SI
system.
Ans. Horse power is the British unit of power. It is the average power which a horse
can develop.
1 H.P. = 746.6 W ≈ 750 W (approx)
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26. Differentiate between work and power.
Ans.
Work
Power
1. Work is the product of force and
displacement in the direction of force.
2. Its SI unit is joule.
1. Power is the rate of doing work.
2. Its SI unit is watt (Js–1)
27. Make the correct choice
Kilowatt is the unit of :
(a) work
(b) power
(c) force
(d) momentum
Ans. (b) is the correct choice :
28. Make the correct choice.
One horse power is equal to :
(a) 647 W
(b) 746 W
(c) 476 W
(d) 764 W
Ans. (b)746 W is the correct choice.
29. Complete the following sentences.
(a) The SI unit of work is ................ and of power is ............................
(b) Kilowatt is the unit of ..........................
(c) Joule is the unit of ............................. .
(d) 1 joule = .................................... ergs.
(e) 1 HP = ..................................... watts.
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Ans.
(a) (i) joule, (ii) watt.
(b) power
(c) work
(d) 107
(e) 746
30.
(a) Define the term energy.
(b)Define the term potential energy. Give four examples of potential energy.
(c)Define the term kinetic energy. Give four examples of kinetic energy.
Ans.
(a) The capacity of doing work is called energy.
(b) The energy possessed by a body, by virtue of its position or
configuration, is called potential energy.
Examples :
(i) A stretched bow and arrow system.
(ii) Wound up spring of a watch.
(iii) Water stored high up in reservoirs.
(iv) A stone lying on the top of the roof.
(c) The energy possessed by a body by virtue of its motion is called kinetic
energy.
Examples :
(i) A running horse
(ii) Speeding car
(iii) Flowing water
(iv) Flying bird.
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31. Derive an expression for potential energy.
Or
Derive an expression for potential energy of a body of mass m, placed at a
height ‘h’ above the earth’s surface.
Ans. Consider a body of mass ‘m’, raised vertically upward through a height ‘h’, to
a new position, against the acceleration due to gravity ‘g’.
∴ Force acting on body = mg.
∴ Work done or energy spent = Force × displacement = mg × h.
But, work done in raising a body to the new position is called potential
energy.
∴
P.E. = mgh.
32. Derive an expression for kinetic energy.
Or
A body of mass m is moving with a velocity v. Derive the expression for the
kinetic energy.
Ans. Consider a body of mass ‘m’ initially moving with a velocity ‘v’ brought to
rest through a distance ‘x’, such that ‘a’ is the constant negative acceleration
acting on the body.
Applying
v2 – u2 = 2aS.
(0)2 – (v)2 = 2ax.
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v2
a= −
2x
∴
∴ Retardation acting on body
⎛ v2 ⎞
v2
= − (a) = − ⎜ −
⎟ = 2x
⎝ 2x ⎠
mv 2
∴ Force acting on the body = m.(–a) =
2x
∴ Total work done in stopping the body
mv 2
×x=
=F×S=
2x
1
2
mv2
But, total work done in stopping the body = K.E.
∴
K.E. =
1
mv2
2
33. (a) State the law of conservation of energy.
(b) How can you demonstrate the law by simple experiment?
Ans. (a) The energy can neither be created, nor destroyed. It may be transformed
from one form to another form, but total energy of the system remains
constant. This is the law of conservation of energy.
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(b) Diagram below shows a freely oscillating pendulum.
In position A, it has max. P.E. and zero K.E.
In position B, it has zero P.E. and max K.E.
In position C, it has max. P.E. and zero K.E.
Thus, experiment clearly proves that the sum total of energy is conserved,
but it can change its form.
34. State the energy changes which occur in following cases.
(a) Burning of coal.
(b) Petrol engine of running motor car.
(c) An electric cell in a circuit.
Ans. (a) The chemical energy of coal, changes into heat energy.
(b) The chemical energy of petrol changes into heat energy. The heat energy
changes into kinetic energy and drives the car.
(c) The chemical energy of cell, changes into electric energy.
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35. (a) Give an example, when work done by a body is zero, even when body
gets displaced from its original position on the application of force.
(b) State the energy changes brought about by the following devices :
(i) Microphone
(ii) Steam engine
(iii)An electric bulb
Ans.
(a) A body revolving around its own axis, such as a working electric fan.
(b) (i) In microphone, the sound energy changes into electric energy.
(ii) In steam engine, the chemical energy of coal changes into heat
energy. The heat energy changes into kinetic energy of the steam
and drives the engine.
(iii) In an electric bulb the electric energy first changes into heat energy.
The heat energy then changes into light energy.
36. A tennis ball and a table tennis ball have same momentum. Which of the two
has more kinetic energy and why?
Ans. Table tennis ball has more kinetic energy.
It is because its mass is very small and hence in order to have same
momentum, as tennis ball, it must have more velocity than the tennis ball and
consequently more kinetic energy.
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37. A man climbs a slope and another walks same distance on level road.
Which of the two expends more energy and why?
Ans. The man moving along the slope has more energy. It is because, while moving
along slope, he gains height, and hence, has extra potential energy.
38. What kind of energy is possessed in the following situations?
(a) A locked up spring of an air gun.
(b) A stone lying on the top of a roof.
(c) A fish moving in water.
(d) A horse running along a level road.
(e) Water stored in the dams.
(f) An electron spinning around the nucleus.
(g) A shooting arrow.
(h) A stone, in a stretched catapult.
Ans. (a) P.E.
(b) P.E.
(c) K.E.
(d) K.E.
(e) P.E.
(f) K.E.
(g) K.E.
(h) P.E.
39. Give one example in each case, when :
(a) Heat energy changes into kinetic energy.
(b) Kinetic energy changes into heat energy.
(c) Sound energy changes into electric energy.
(d) Electric energy changes into sound energy.
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(e) Light energy changes into chemical energy.
(f) Chemical energy changes into light energy.
(g) Electric energy changes into mechanical energy.
(h) Mechanical energy changes into electric energy.
(i) Potential energy changes into electric energy.
Ans. (a) In the automobile engine, heat energy changes into kinetic energy.
(b) When two stones are rubbed against each other, the kinetic energy changes
into heat energy.
(c) When we speak in front of microphone, the sound energy changes into
electric energy.
(d) When a loudspeaker works, the electric energy changes into sound energy.
(e) During photosynthesis, the light energy changes into chemical energy.
(f) During burning of magnesium, the chemical energy changes into light
energy.
(g) In an electric motor, the electric energy changes into mechanical energy.
(h) In an electric generator, the mechanical energy changes into electric
energy.
(i) The P.E. of stored water in the dams, changes into the electric energy in the
generators.
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40. State the energy changes which take place in the following cases :
(a) A bulb glows, when torch light is switched on.
(b) A car moves up a hill.
(c) A toy car spring is wound and the car is made to run on level floor.
(d) Water stored in the dams rotates the turbine connected to a dynamo.
(e) An air gun is loaded and then fired.
(f) A magnesium ribbon burns in air.
(g) A stone projected vertically upwards, returns back to the thrower.
(h) Water freezes in the freezing chamber of a fridge.
(i) Photographic film is exposed to the sunlight.
(j) Food is digested by animals.
Ans. (a) The chemical energy of cell changes into electric energy. The electric
energy first changes to heat and finally into light energy.
(b) The K.E. of moving car changes into potential energy.
(c) The mechanical energy changes into potential energy during winding. The
potential energy is then released in the form of kinetic energy.
(d) P.E. of stored water changes into K.E. of the flowing water. The K.E. of the
flowing water rotates the turbine. The turbine in turn rotates the coil of the
generator and changes into electric energy.
(e) During loading, the mechanical energy changes into potential energy of the
spring. On firing, the P.E. of the spring changes into K.E. of the bullet.
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(f) During burning of magnesium, chemical energy changes into heat energy
and light energy.
(g) The kinetic energy changes into the potential energy when stone rises up.
The potential energy then changes back to the kinetic energy, when stone
falls down.
(h) The kinetic energy of water molecules is released in the form of heat
energy.
(i) The light energy changes into the chemical energy.
(j) The chemical energy in the food changes into the heat energy.
41. What physical quantity does the electron volt (eV) measure? How is it related
to SI unit of energy?
Ans. Electron volt measures the energy possessed by electrons moving in an electric
field.
1 eV = 1.6 × 10–19 J.
42. Complete the statement.
1 joule = ........................... calorie.
Ans. 1 joule = 0.238 calorie.
43. Name the physical quantity which is measured in calorie. How is it related to
SI unit of that quantity?
Ans. Physical quantity measured in calorie is heat energy. 1 calorie = 4.2 J.
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44. How much energy per second does a 40 watt bulb require?
Ans. It requires energy at a rate of 40 Joule per second.
45. What are two forms of mechanical energy?
Ans. (i) Kinetic energy (ii) Potential energy
46. What is meant by gravitational potential energy.
Ans. The potential energy possessed by a body at some height above the ground
level, is called gravitational potential energy.
47. How is gravitational potential energy measured quantitatively?
Ans. It is measured as the amount of work done in lifting a body to a certain height,
against the force of gravity.
48. Name the form of energy which a body may possess, even when it is not in
motion. Give an example in support of your anwer.
Ans. The body may possess potential energy. For example, a stone lying on the roof
of a house is not in motion, but possesses potential energy.
49. State the work-energy theorem.
Ans. The work done by a force is equal to the increase in its kinetic energy.
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50. 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 to v. How much work is being
done by the force?
Ans. Work done by the force
= Change in the kinetic energy.
= Final kinetic energy – Initial kinetic energy
=
1
1
1
mv2 – mu2 = m(v2 – u2)
2
2
2
51. A light mass and a heavy mass have equal momentum. Which will have more
kinetic energy? Explain.
Ans. We know, momentum = mass × velocity
p = mv
Squaring both sides, we get p2 = m2v2
v2 =
Also,
kinetic energy =
p2
…(i)
m2
1
mv2
2
…(ii)
Substituting the value of (i) in (ii)
1
p2
p2
Kinetic energy = m × 2 =
2
2m
m
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p2
In the expression
, for the same momentum, if m is less, then magnitude
2m
of kinetic energy will be more.
Thus, the lighter body will have more kinetic energy than the heavier body for
the same momentum.
52. Name three forms of kinetic energy and give one example each?
Ans.
1. Translational kinetic energy : This energy is possessed by a moving car
in a straight line.
2. Rotational kinetic energy : This energy is possessed by a wheel rotating
about its axis.
3. Vibrational kinetic energy : This energy is possessed by a vibrating string
of a guitar or violin.
53. What do you understand by conservation of mechanical energy?
Ans. Whenever there is an interchange between the potential energy and the kinetic
energy, the total mechanical energy (sum total of kinetic energy and potential
energy) remains constant, provided no frictional forces are acting.
54. Name two examples in which mechanical energy of a system remains
constant.
Ans.
1. When a pendulum is oscillating freely about its mean position.
2. When a stone projected vertically upward it returns back to the thrower.
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55. A body falls freely under the action of gravity from the position of rest.
Name the kind of energy it possessed (i) at the point from where it fell,
(ii) while falling, (iii) on reaching ground.
Ans.
(i) It has only potential energy.
(ii) It has both potential energy and kinetic energy.
(iii) It has only kinetic energy.
56. A body is thrown vertically upward. Its velocity keeps on decreasing. What
happens to its kinetic energy as its velocity becomes zero.
Ans. The kinetic energy of the body completely changes into potential energy at the
highest point.
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Numerical Problems.
1. Two bodies of equal masses are moving with velocities v and 3 v respectively.
Find the ratio of their kinetic energies.
Ans. K.E. of Ist body : K.E. of 2nd body =
1
1
m(v) 2 : m(3v) 2 = 1 : 9.
2
2
2. The speed of motor bike changes from 5 ms–1 to 20 ms–1. Find the ratio of
initial and final kinetic energies.
Ans. Initial K.E. : Final K.E. =
1
1
m(5)2 : m(20)2
2
2
= 25 : 400 = 1 : 16.
3. A ball X is kept at a height of h. Another, similar ball is kept at a height of 5h.
What is the ratio of their potential energies?
Ans. P.E. of Ist ball : P.E. of second ball = mg(h) : mg (5h) = h : 5h = 1 : 5.
4. A 200 g ball is thrown vertically upward with an initial velocity of 30 ms–1.
(a) Draw a velocity-time graph for the motion of the ball. (acceleration due to
gravity is 10 ms–2).
(b) How long will the ball take to reach highest point?
(c) What will be the kinetic energy of ball, when it returns to starting point?
(Ignore air resistance).
(d) What will be the potential energy of ball at highest point?
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Ans. m = 200 g = 0.200 kg; u = 30 ms–1
(a) Shown in the diagram below :
(b) 3 seconds
(c) K.E. =
=
1
mv2
2
1
× 0.200 × (30)2 = 90 J.
2
(d) P.E. = K.E. [By the law of conservation of energy]
= 90 J.
5. A water pump raises 50 kg of water, through a height of 25 m in 5 seconds.
Calculate the power at which pump works.
(Take g = 10 Nkg–1).
Ans. m = 50 kg;
h = 25 m;
t = 5 s;
g = 10 Nkg–1;
P=?
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Work done by the pump = P.E = mgh = 50 × 10 × 25
= 12500 J
∴ Power of pump =
w 12500 J
=
= 2500 W.
t
5s
6. A metal ball of mass 2 kg is allowed to drop freely from rest, from a height of
5 m above ground.
(i) Calculate P.E. possessed by ball.
(ii) What is the kinetic energy of ball, just before hitting the ground?
(iii) What happens to mechanical energy, after the ball hits ground and comes
to rest?
Ans. m = 2 kg;
h = 5 m;
g = 10 ms–2
(i) P.E. = mgh = 2 × 10 × 5 = 100 J.
(ii) K.E. = P.E. = 100J. [By the law of conservation of energy].
(iii) The kinetic energy is dissipated in the form of sound energy and heat energy.
7. A boy of mass 40 kg, runs upstairs and reaches 8 m high first floor, in 5 s.
Calculate : (i) Force of gravity acting on boy, (ii) Work done by him against
[g = 10 ms–2]
gravity, (iii) Power developed by boy.
Ans. Mass of the boy = 40 kg;
Vertical height through which boy climbs up = 8 m;
Time = 5 seconds
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(i) Force of gravity acting on boy
= mg = 40 kg × 10 ms–2 = 400 N.
(ii) Work done by boy = Force × displacement
= 400 N × 8 m = 3200 J.
(iii) Power developed by boy =
w 3200 J
=
t
5s
= 640 W.
8. The world record for weight lifting is held by Sergei Didyk of U.S.S.R. He lifted
261 kg to a height of 2.3 m in 4 s. Find : (i) weight lifted by Didyk,
(ii) work done by him, (iii) power developed by him.
[g = 10 ms–2]
Ans. m = 261 kg;
h = 2.3 m;
t = 4 s;
g = 10 ms–2
(i) Weight lifted = mg = 261 × 10 = 2610 N.
(ii) Work done = F × h
= 2610 × 2.3 = 6003 J.
(iii) Power developed =
w 6003 J
=
t
4
= 1500.75 W.
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9. A boy weighing 350 N, runs up a flight of 30 steps, each 20 cm high, in 5
seconds. Calculate the power expended. [g = 10 ms–2]
Ans.Weight (mg) = 350 N; Time = 5 s;
g = 10 ms–2;
Distance = 30 × 20 cm = 600 cm = 6 m
Work done by the boy = Force × Displacement
= 350 N × 6 m = 2100 J
∴
Power expended =
2100 J
= 420 W.
5s
10. A person of mass 60 kg, runs up a flight of 30 steps, each measuring 20 cm in
height, in 15 s. Calculate the power developed by a person.
[Take g = 10 ms–2]
Ans. Mass of person = 60 kg;
Height = 30 × 20 = 600 cm = 6 m; Time = 15 s;
P = ?; g = 10 ms–2.
Work done by person = P.E. = mgh
= 60 × 10 × 6 = 3600 J
Power developed =
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W 3600 J
=
= 240 W.
15 s
t
25
Question Bank
11. A body of mass 1 kg is thrown vertically upward with an initial speed of
5 ms–1. What is the magnitude of force due to gravity acting on body, when it is
at the highest point? What is the maximum height attained by the body?
[g = 10 ms–2]
Ans. m = 1 kg; u = 5 ms–1; v = 0;
h = ?;
g = 10 ms–1; F = ?
Magnitude of force = mg = 1 kg × 10 ms–2
= 10 N.
Applying,
v2 – u2 = –2gh
(0) 2 – (5)2 = –2 × 10 × h
∴
h=
–25
= 1.25 m.
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12. A block of mass 30 kg is pulled up along a slope, as shown in the diagram with a
constant speed, by applying a force of 200 N parallel to slope. A and B are initial
and final positions of the block.
(i) Calculate the work done by the force in moving the block from A to B.
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(ii) Calculate P.E. gained by block.[g = 10 ms–2]
Ans. (i) Work done in moving the block from A to B = 200 N × 3 m = 600 J.
(ii) Gain in potential energy = mgh = 30 × 10 × 1.5 = 450 J.
13. A work of 1000 J is done on a body in 4 s, such that displacement of 20 m is
caused. Calculate :
(a) Force, (b) Power.
Ans. W = 1000 J;
t = 4 s;
S = 20 m;
F=?
P=?
(a) Force = Work ÷ Displacement
= 1000 J ÷ 20 m = 50 N.
(b) Power = Work ÷ Time = 1000 J ÷ 4 s = 250 W.
14. An engine of an automobile of power 200 W, works for 4 s. If the force developed
by engine is 100 N, calculate the maximum displacement it can cause.
Ans. P = 200 W;
t = 4 s;
F = 100 N;
S=?
W=P×t
= 200 W × 4 s = 800 J
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Work = Force × Displacement
∴
800 J = 100 N × Displacement
∴ Displacement = 8 m.
15. Calculate the horse power of the motor of an elevator, which can carry 10
persons of average mass 60 kg, through a vertical height of 20 m in 30 s.
[Take g = 10 Nkg–1]
Ans. Mass of ten persons = 60 kg × 10 = 600 kg
Height = 20 m;
g = 10 Nkg–1;
t = 30 s.
∴
Work done = mgh
= 600 kg × 10 N kg–1 × 20 m = 120000 J.
∴ Power developed = Work ÷ Time
= 120000 J ÷ 30 s = 4000 W.
∴
Power in H.P. = 4000 ÷ 750 = 5.33 HP. [∵ 1 H.P ≈ 750 w (Approx)]
16. How long should an electric motor of power 2 H.P. operates, so as to pump
5 m3 of water from a depth of 15 m.
[Take g = 10 Nkg–1]
Ans. Mass of water = Volume × Density
= 5 m3 × 1000 kg m–3 = 5000 kg
Height = 15 m;
t = ?;
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Question Bank
g = 10 Nkg–1;
P = 2 HP = 2 × 750 = 1500 W.
Work done in operating electric motor
= P × t = 1500 W × t.
Work done in raising water
= mgh = 5000 × 10 × 15
= 750000 J
∴
1500 W × t = 750000 J
t = 500 s.
17. An electric pump is 60% efficient and is rated 2 H.P. Calculate the maximum amount
of water, it can lift through a height of 5 m in 40 s.
[Take g = 10 ms–2 and 1 HP = 750 W]
Ans. Power = 2 H.P = 2 × 750 = 1500 W;
Efficiency = 60%.
∴ Useful power= 1500 ×
60
= 900 W.
100
∴ The work done by motor in 40 s
= 900 W × 40 s = 36000 J.
Also, work done by motor in lifting, water
= mgh = m × 10 × 5 = 50 m
∴
or,
50 m = 36000
Maximum amount of water lifted,
m=
Class-X
36000
= 720 kg.
50
29
Question Bank
18. A compressed spring is held near a small toy car of mass 0.15 kg. On release of
spring, the toy car moves forward with a velocity of 10 ms–1. Find the potential
energy of the spring.
Ans. K.E. of toy car
1
1
mv2 =
× 0.15 kg × (10 ms–1)2 = 7.5 J
2
2
∴ Potential energy of the spring
= K.E. of toy car = 7.5 J.
19. 5 × 1012 alpha particles pass through a point in 40 s. If the velocity of these
particles is 2 × 107 ms–1, calculate : (a) energy possessed by these particles,
(b) power developed.
[Mass of an alpha particle is 6.64 × 10–27 kg].
Ans. K.E. of one alpha particle
=
1
2
1
mv2 = × 6.64 × 10–27× (2 × 107)2
2
= 3.32 × 4 × 10–13 J
= 13.28 × 10–13J
(a) K.E. of 5 × 1012 alpha particles = 5 × 1012 × 13.28 × 10–13J = 6.64 J.
(b) Power developed
=
Class-X
Work 6.64 J
=
= 0.166 W.
Time
40 s
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Question Bank
20. A body ‘A’ of mass 20 kg is moving with a velocity of 1 ms–1. Another body ‘B’ of
mass 1 kg is moving with a velocity of 20 ms–1. Find the ratio of kinetic energies of
A and B.
1
mv2
2
Ans. K.E. of A ‘E1’=
1
× 20 kg × (1 ms–1)2 = 10 J
2
=
K.E. of B (E2) =
=
1
2
mv2
1
× 1 kg × (20 ms–1)2
2
= 200 J
∴
E1 : E2 = 10 : 200 = 1 : 20.
21. A bullet of mass 0.2 kg is moving with a velocity of 200 ms–1 and strikes a stationary
wooden target of mass 5 kg. If all the energy is transferred to wooden target, calculate
the velocity of target.
Ans.
1
mv2
2
K.E. of bullet =
=
1
× 0.2 kg × (200 ms–1)2
2
= 4000 J
K.E. of wood =
1
2
mv2 =
1
2
× 5 kg × v2
= 2.5 v2
2.5 v2 = 4000
∴
4000
= 1600
2.5
v = 1600 = 40 ms–1.
v2 =
∴
Class-X
31
Question Bank
22. A bullet of mass 0.05 kg strikes a wooden target 0.2 m thick, with a velocity of
300 ms–1 and emerges out with a velocity of 100 ms–1. Calculate : (i) loss in
K.E., (ii) average force of friction of wood.
Ans. (i)
Loss in K.E. =
=
(
1
m v 21 – v 22
2
)
1
× 0.05 × [(300) 2 – 100) 2 ]
2
= 0.025 × 80000 = 2000 J.
Loss in K.E. = Retarding force × distance
(ii)
2000 J = Retarding force × 0.2 m
∴ Retarding force =
2000 J
= 10000 N.
0.2 m
23. 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) power developed by him.
[Take g = 10 Nkg–1]
Ans. (i)Force = 200 kgf = 200 kg × 10 Nkg–1
= 2000 N
Height = 2.5 m
∴ Work done = P.E. = mgh
= 2000 N × 2.5 m = 5000 J
(ii)
Class-X
Power =
Work 5000 J
=
= 1000 W
Time
5s
32
Question Bank
24. A body, when acted upon by a force of 10 kgf gets displaced by 0.5 m. Calculate
the work done by the force, when displacement is : (i) in the direction of force
(ii) at an angle of 60° with the force and (iii) at normal to the force.
Ans.
(i) Work done = F × S = 10 kgf × 0.5 m
= 100 N × 0.5 m = 50 J
(ii) Work done = F cos θ × S
= 100 N × cos 60° × 0.5 m
= 100 N ×
1
× 0.5 m
2
= 25 J
(iii) Work done = F cos θ × S
= 100 N × cos 90° × 0.5 m
= 100 N × 0 × 0.5 m = zero.
25. A boy of mass 40 kg runs upstairs and reaches 8 m high first floor in 5 s.
Calculate :
(i) Force of gravity acting on the boy.
(ii) The work done by him against the gravity.
[Take g = 10 ms–2]
Ans.
(i) Force acting on the boy = mg
= 40 kg × 10 ms–2
= 400 N
(ii) Work done by the boy = F × S
= 400 N × 8 m
= 3200 J
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33
Question Bank
26. A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute.
Calculate the work done and power spent.
[g = 10 ms–2]
Ans. Height covered by the boy = 30 × 20 cm
= 600 cm = 6 m
Work done by the boy = F × S = 350 N × 6 m
= 2100 J
Power spent by the boy =
W 2100 J
=
= 35 W
t
60 s
27. It takes 20 s for A to climb up the stairs, while B does in 15 s. Compare the :
(i) work done and
(ii) power developed by A and B. Assume boys have same mass.
Ans.
(i) As the boys A and B have same mass and climb through same height
against the gravity, therefore, the work done (W), by them is same.
Work of A : Work of B = W : W = 1 : 1
(ii) Power of A (PA) =
W
W
=
20 s
t
Power of B (PB) =
W
W
=
15 s
t
PA : PB =
Class-X
W
W
:
=3:4
20 s 15 s
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Question Bank
28. A machine raises a load of 750 N through a height of 16 m in 5 s. Calculate :
(i) Work done by the machine
(ii) power at which machine works
Ans. (i) Work done by the machine
= F × S = 750 N × 16 m = 12000 J
(ii) Power of machine =
W 12000 J
=
= 2400 W
5s
t
29. A boy of mass 40 kg runs up a flight of 15 steps, each 15 cm high in 10 s. Find
the work done and power developed by him. [Take g = 10 N kg–1.]
Ans. Force acting on the boy (F) = mg
= 40 kg × 10 Nkg–1
= 400 N
Distance covered by the boy (h) = 15 × 15 cm
= 225 cm
= 2.25 m
Work done by the boy (W) = F × h
= 400 N × 2.25 m
= 900 J
Power developed by the boy (P) =
W 900 J
=
t
10 s
= 90 W
Class-X
35
Question Bank
30. A water pump raises 50 litres of water through a height of 25 m in 5 s. Calculate
the power which pump supplies.
[Take g = 10 Nkg–1 and density of water
= 1000 kgm–3]
Ans. Volume of water = 50 litres =
50
m3 = 0.05 m3.
1000
Mass of water = V × d
= 0.05 m3 × 1000 kgm–3 = 50 kg
Work done in lifting water = mgh
= 50 kg × 10 Nkg–1 × 25 m = 12500 J
W 12500 J
=
Power supplied by pump =
= 2500 W
t
5s
31. A man raises a box of 50 kg mass to a height of 2 m in 2 minutes while another
man raises the same box to the same height in 5 minutes. Compare : (i) work
done and (ii) the power developed by them.
Ans. (i) As both men raise the box of same mass through same height against the
force of gravity, therefore, they do same amount of work (W).
Work done by 1st man : work done by 2nd man = W : W = 1 : 1
W
W
=
t
2 min
W
W
=
Power of 2nd man (P2) =
t
5 min
W
W
:
= 5: 2
∴ P1 : P2 =
2 min 5 min
(ii) Power of 1st man (P1) =
Class-X
36
Question Bank
32. A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. Calculate :
(a) work done by the pump.
(b) the power at which pump works
(c) the power rating of pump if its efficiency is 40%.
[Take g = 10 ms–2]
Ans. (a) Work done by pump = PE = mgh
= 500 kg × 10 ms–2 × 80 m = 400000 J
W
4 ×105 J
(b) Power of pump =
=
t
10 s
= 4 × 104 W = 40 kW.
(c) Power rating (power input) =
Useful power 40 kW
= 100 kW
=
40/100
Efficiency
33. An ox can apply a maximum force of 1000 N. It is taking part in a cart race and
is able to pull a cart at a constant speed of 30 ms–1, while making its best effort.
Calculate the power developed by the ox.
Ans. Power of ox = Force × constant speed
= 1000 N × 30 ms–1
= 30,000 W = 30 kW
34. If the power of a motor is 40 kW, at what speed can it raise a load of 20,000 N.
Ans.Power of motor = 40 kW = 40,000 W
Also, power of motor = Force × Constant speed
Constant speed =
Power of motor 40,000 W
=
Force
20,000 N
= 2 ms–1.
Class-X
37
Question Bank
35. Find the kinetic energy of a body of mass 1 kg, moving with a uniform velocity
of 10 ms–1.
Ans. K.E. =
1
mv2
2
=
1
(1 kg)(10 ms–1)2 = 50 J
2
36. A ball of mass 0.5 kg slows down from a speed of 5 ms–1 to that 3 ms–1.
Calculate the change in kinetic energy of the ball.
1
mv2
2
Ans. Initial K.E. of the ball =
=
1
2
× 0.5 kg (5 ms–1)2
= 6.25 J
Final K.E. of the ball =
=
1
mv2
2
1
2
× 0.5 kg (3 ms–2)2
= 2.25 J
∴ Decrease in kinetic energy of the ball
= (6.25 – 2.25) J
= 4.0 J
Class-X
38
Question Bank
37. A canon ball of mass 500 g is fired with a speed of 15 ms–1. Find its kinetic
energy and momentum.
Ans. Kinetic energy =
=
1
mv2
2
1 ⎛ 500
⎞
kg ⎟ (15 ms –1 ) 2
×⎜
2 ⎝ 1000 ⎠
= 56.25 J
Momentum = mv =
500
kg ×15 ms –1
1000
= 7.5 kg ms–1
38. A bullet of mass 50 g 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) average retarding force offered by target.
Ans. (i) Kinetic energy of the bullet
mv2
=
1
=
1 ⎛ 50
⎞
kg ⎟ (500 ms–1)2 = 6250 J.
⎜⎝
2 1000 ⎠
2
(ii) Kinetic energy of the bullet
= Work done in stopping bullet
= 6250 J
Class-X
39
Question Bank
Also, work done in stopping bullet
= Retarding force × Displacement
⎛ 10 m ⎞
6250 J = Retarding force × ⎜
⎝ 100 ⎟⎠
Retarding force =
6250 × 100 J
10 m
= 62500 N
39. A truck weighing 1000 kgf changes its speed from 36 kmh–1 to 72 kmh–1 in
2 minutes. Calculate the work done by the engine and its power. (g = 10 ms–2)
Ans.Weight of the truck = 1000 kgf
∴ Mass of the truck = 1000 kg
Initial speed of truck = 36 kmh–1 = 10 ms–1
Final speed of truck = 72 kmh–1 = 20 ms–1
∴ Initial K.E. of the truck
=
1
1
mv2 = × 1000 kg × (10 ms–1)2
2
2
= 50000 J
Final K.E. of the truck
Class-X
=
1
mv2
2
=
1
× 1000 kg (20 ms–1)2
2
40
Question Bank
= 200000 J
∴ Work done by engine = change in K.E.
= (200,000 – 50,000)J = 150,000 J
Power of engine =
W 150, 000 J
=
= 1250 W.
t
2 × 60 s
40. A body of mass 60 kg has momentum 3000 kg ms–1. Calculate the speed and
kinetic energy of the body.
Momentum = Mass × Speed
Ans.
3000 kgms–1= 60 kg × speed
3000 kgms –1
= 50 ms–1
Speed =
60 kg
1
1
mv2 = × (60 kg)(50 ms–1)2
2
2
= 75,000 J
41. How much work will have to be done on a ball of mass 50 g to give a momentum
Kinetic energy =
of 500 g cm s–1?
Momentum = Mass × Speed
500 g cm s–1 = 50 g × Speed
500 gcms –1
Speed =
50 g
= 10 cms–1 = 0.1 ms–1
Ans.
Work done = KE of the ball =
=
Class-X
1
mv2
2
1 ⎛ 50
⎞
kg ⎟ (0.1 ms–1)2
×⎜
2 ⎝ 1000 ⎠
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Question Bank
= 2.5 × 10–4 J.
42. A ball of mass 0.20 kg is thrown vertically upward with an initial velocity of
20 ms–1. Calculate the maximum potential energy it gains as it goes up.
Ans. Maximum potential energy gained
= Maximum K.E. at the starting point
=
1
1
mv2 = (0.20 kg)(20 ms–1)2
2
2
= 40 J
43. A metal ball of mass 2 kg is allowed to fall freely from rest at a height of 5 m
above the ground.
(i) Calculate the potential energy possessed by the ball, when initially at rest.
(ii) What is the kinetic energy of the ball just before it hits the ground.
(iii) What happens to the mechanical energy after the ball hits the ground and
comes to rest.
Ans. (i) Initial potential energy = mgh
= (2 kg) × (10 ms–2) × (5 m)
= 100 J
(ii) By the law of conservation of energy :
K.E. just before hitting ground
= Initial potential energy
= 100 J
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Question Bank
(iii) The mechanical energy on hitting the ground is dissipated in the form of heat
energy and sound energy.
44. The diagram given below shows a ski jump. A skier weighing 60 kgf stands at A,
at the top of ski jump. He moves from A to B and takes off for his jump at B.
(a) Calculate the change in gravitational potential energy of the skier between A
and B.
(b) If 75% of the energy in part (a) becomes kinetic energy at B, Calculate the
speed at which skier arrives at B. [Take g = 10 ms–2]
Ans. (a) Weight of the skier = 60 kgf
∴ Mass of the skier = 60 kg
Initial gravitational P.E. of skier = mgh
= 60 kg × 10 ms–2 × 75 m
= 45000 J
Final gravitational P.E. of skier
= 60 kg × 10 ms–2 × 15 m
= 9000 J
Decrease in the gravitational P.E. of skier
Class-X
43
Question Bank
= (45000 – 9000) J = 36000 J
(b) 75% of the P.E. of skier
=
75
× 36000 J
100
= 27000 J.
Now 27000 J of P.E. is utilised in imparting kinetic energy.
Kinetic energy of the skier = 27000 J.
Also kinetic energy of the skier
=
1
mv2
2
=
1
× 60 kg × v2
2
1
× 60 kg × v2 = 27000 J
2
v2 =
27000 J
30 kg
= 900 m2s–2
V = 900 m 2s –2
= 30 ms–1.
Class-X
44
Question Bank
45. A hydroelectric power station takes its water from a lake whose level is 50 m
above the turbine. Assuming overall efficiency of 40%, calculate the mass of
water which must flow through the turbine each second to produce power output
of 1 MW.
Ans. Useful power output = 1 MW
Actual power output =
100
× 1 MW
40
= 2.5 MW.
Actual work done by water falling from height in 1s
2.5 MJ = 25,00000 J
Also work done by falling water
= PE = mgh
= m × 10 ms–2 × 50 m
m × 10 ms–2 × 50 m = 2500000 J
Mass of water falling per second =
2500000 J
500 m 2s –2
= 5000 kg.
Class-X
45
Question Bank