(Chapter – 11) (Work and Energy)
1 Mark each
Question 1:
Name two devices which convert electrical energy into sound.
[CBSE 2012]
Answer 1:
Loudspeaker and microphone.
Question 2:
Identify the kind of energy possessed by a running athlete.
[CBSE 2011]
Answer 2:
Kinetic energy.
Question 3:
The potential energy of a free falling object decreases progressively. Does this violate the law of conservation of energy? Why?
[CBSE 2011]
Answer 3:
Decrease in P.E. = Increase in K.E. So there is no loss in the energy by the object. There is only the transformation of the energy.
Question 4:
When a ball is thrown vertically upwards, its velocity goes on decreasing. What happens to its potential energy as its velocity becomes zero?
[CBSE 2011]
Answer 4:
According to the law of conservation, energy can neither be created nor be destroyed but can transform from one form to another.
Therefore, K.E. Lost by the ball = P.E. gained by the ball
So, as velocity of ball becomes zero then K.E. would be zero and the P.E. would be maximum.
Question 5:
If the heart works 60 joules in one minute, what is its power?
[CBSE 2011]
Answer 5: Power = Work Time done
= 60 60 J
s = 1W. [ 1 Watt, W = J/s]
Question 6:
A coolie is walking on a railway platform with a load of 30 kg on his head. How much work is done by coolie? [CBSE 2011]
Answer 6:
Coolie is carrying a load on his head so force exerted by it under gravity and displacement of the load makes 900 angle.
W = 0 [ ∵ Force is perpendicular to the displacement ]
Therefore, Work done by the coolie is zero.
Question 7:
A horse of mass 210 kg and a dog of mass 25 kg are running at the same speed. Which of the two possesses more kinetic energy? How?
[CBSE 2011]
Answer 7: Kinetic energy, K.E. = 12mv2 If the velocity is constant, Kinetic energy is directly proportional to mass. So as mass increases, kinetic energy also increases. Here, mass of horse is greater than that of dog, so kinetic energy of the horse is greater than dog when speed is constant.
Question 8:
What will cause greater change in kinetic energy of a body? Changing its mass or changing its velocity.
[CBSE 2010]
Answer 8:
As KE = 12mv2 therefore,
K.E.∝ m
K.E.∝ v2
Therefore, the change in velocity of a body will cause greater change in kinetic energy.
2 Marks each
Question 1:
Identify and state the type of transformation of energy in the following cases:
(a) When coal is burnt.
(b) In a thermal power plant
[CBSE 2014]
Answer 1:
(a) Chemical energy stored in coal is converted to heat energy.
(b) Chemical energy of fuel is first converted into heat energy than to kinetic energy and finally to electrical energy.
Question 2:
(a) What is meant by potential energy of a body?
(b) A body of mass m is raised to a vertical height h through two different paths A and B. What will be the potential energy of the body in the two cases? Give reason for your answer.
[CBSE 2014]
Answer 2:
(a) Potential energy is the energy possessed by a body due to its position and shape.
(b) The potential energy in both the cases is equal.
P.E. = mgh
It is independent of the path followed for attaining the height.
Question 3:
Calculate the total energy consumed in the month of November in a household in which four devices of power 500 W each are used daily for 10 h.
[CBSE 2013]
Answer 3:
Given,
Power, P = 500 W;
Time, t = (4 × 10 × 30) h = 1200 h
Total energy consumed in the month of November
= P × t
= 500 × 1200
= 600000 Wh = 600 kWh [ ∵ 1 kWh =1000W]
= 600 unit. [ ∵ 1 kWh =1 unit ]
Question 4:
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
[CBSE 2012]
Answer 4:
Work done = Force × displacement
= 140 × 15
= 2100 J = 2.1 kJ
Question 5:
A body of mass 2 kg is thrown up with a speed of 25 m/s. Find its maximum potential energy.
[CBSE 2012]
Answer 5:
Maximum kinetic energy of the body = Maximum potential energy of the body.
Given,
Mass of the body, m = 2 kg
Speed, v = 25 m/s
Therefore, = 12 × 2 × 25 K.E. × 25
= 12mv2
= 625 J.
Question 6:
Define work. How is work done measured?
[CBSE 2012]
Answer 6:
Work is said to be done whenever a force acts on a body causing its motion in the direction of the force.
Work done = Force × Displacement
For example, when a force F acts on a body it moves through a distance‘s’ in the direction of the force.
Therefore, the work done, W = F × Question 7:
Is there any work done on a body in uniform circular motion? Justify your answer.
[CBSE 2012]
Answer 7:
If a body moves in uniform circular motion, the centripetal force ‘F’ acts along the radius towards the Centre and the displacement ‘s’ acts along the tangent to the circle. Thus, the angle between ‘F’ and‘s’ is 90°. Therefore,
W = 0 [ ∵ Force is perpendicular to the displacement ]
Since W= 0, no work is done by the body in uniform circular motion.
Question 8:
Given below is a few situations, study them and state in which of the given cases work is said to be done. Give reason for your answer.
(a) A person pushing hard a huge rock but the rock does not move.
(b) A bullock pulling a cart up to 1 km on road.
(c) A girl pulling a trolley for about 2m distance.
(d) A person standing with a heavy bag on his head.
[CBSE 2012]
Answer 8:
(a) Work done = 0,
Reason- As the rock has not moved so displacement is zero.
(b) Work done = Positive
Reason – As the motion is in the direction of force applied.
(c) Work done = Positive
Reason – As the trolley is moving in the same direction in which force is applied.
(d) Work done = 0
Reason – As there is no movement in the position of the bag, therefore displacement is zero.
Question 9:
What is power? What is its SI unit?
[CBSE 2010]
Answer 9:
The rate at which work is done is called power.
Power = work done / time. Its SI unit is Watt.
3 Marks each
Question 1:
When a force retards the motion of a body, what is the nature of work done by the force? State reason. List two examples of such a situation.
[CBSE 2014, 2013]
Answer 1:
The work done by the force is negative because this force acts opposite to the direction of movement of the body, thereby hindering the motion.
Example:
1. Work done by the force of friction acting opposite to the direction of football.
2. Work done in Tug of war by losing team.
Question 2:
A lift carries a maximum weight of 2400 N to a height of 10 m with constant speed of 2 m/s. Find out the power and work done. [CBSE 2013]
Answer 2:
Given,
Force, F= 2400N
Height, h = 10 m
Speed, v = 2 m/s
Therefore, Work done = F × s = 2400 × 10 = 24000 J
And, Power = F × v = 2400 × 2 = 4800 W
Question 3:
What is meant by work done by a force? When do we call work done to be positive or negative? In a game of ‘tug of war’ state the type of work done by the winning and the losing teams.
[CBSE 2013]
Answer 3:
When force applied on a body causes the body to move then the work is said to be done by the force.
Positive work done: when the body is displaced in the same direction that of the force then work done is positive. That is the angle between the force applied and displacement is 0o.
Negative work done: when the body is displaced in the direction opposite to that of the force then work done is negative. Here the angle between the force applied and displacement is 180o.
In the game of ‘tug of war’ work done by the winning team is positive and the work done by the losing team is negative.
Question 4:
Four persons jointly lift a 250 kg box to a height of 1 m and hold it.
(a) Calculate the work done by the persons in lifting the box.
(b) How much work is done for just holding the box?
(c) Why do they get tired while holding it? (g = 10 m/s2).
[CBSE 2012]
Answer 4:
(a) Given,
Force, F = 250 × 10 = 2500 N
Displacement, s = 1 m
Work done = F × s = 2500 × 1 = 2500 J
(b) As there is no displacement in just holding the box, so no work is done i.e. Work done is zero.
(c) To hold the box men are applying force against the gravity, which involves their muscles to work, hence they get tired holding it.
Question 5:
(a) Name the type of energy possessed by a freely falling body of mass ‘m’ at the highest point 'h'.
(b) Define the energy.
(c) A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
[CBSE 2012]
Answer 5:
(a) Potential energy, P.E. is the energy possessed by a body of mass, m after reaching a height, h. it is given by, P.E. = mgh
(b) The capacity to do work is known as Energy.
(c) K.E.of the object is converted into heat and sound energy.
Question 6:
Two masses m and 2 m are dropped from heights h and 2h. On reaching the ground, which body will have a greater kinetic energy and why?
[CBSE 2012]
Answer 6:
K.E. of first mass, m = P.E. lost by mass m = mgh
K.E. of second mass, 2 m = P.E. lost by mass 2 m
= 2 m × g × 2h = 4 mgh
Therefore, mass 2 m will have 4 times greater kinetic energy on reaching the ground.
Question 7:
Water is falling on the blades of a turbine at the rate of 8 × 102 kg per minute, height of fall is 50 m. Calculate the power given to turbine. (g = 10 m/s2).
[CBSE 2012]
Answer 7:
Given,
Height of fall, h = 50 m
Acceleration due to gravity, g = 10 m/s2
Mass of water, m = 8 × 102 kg
Time taken, t = 1 min = 60 s
Therefore, the power given to turbine = Energy
Time = mgh
t = 8×10260
×10×50
= 6.67 × 103 W.
Question 8:
A force applied on a body of mass 4 kg for 5 s changes its velocity from 10 m/s to 20 m/s. Find the power required.
[CBSE 2010]
Answer 8:
Given,
Mass of the body, m = 4kg,
Time taken, t = 5 s,
Initial velocity, u = 10 m/s,
Final velocity, v = 20 m/s,
P =?
Power, P
Change Time taken in K.E.
=
12m(v2 t − u2)
=
12 × 4 × (202 − 102) 5
=120 J/s
The power required is 120 J/s.
5 Marks each
Question 1:
75 kg missile is dropped downwards from an air plane, and has a speed of 60 m/s at an altitude of 850 m above the ground. Determine:
(a) The K.E. possessed by the missile at 850 m.
(b) The P.E. possessed by the missile at 850 m.
(c) The total mechanical energy possessed by the missile.
(d) The K.E. and velocity with which it strikes the ground. [CBSE 2013]
Answer 1:
(a) Given:
Mass of the missile, m = 75 kg and Speed, v = 60 m/s
Therefore, K.E. = 12mv2 = 12 × 75 × 60 × 60 J = 13500 J
(b) Given,
Mass of the missile, m = 75 kg
Acceleration due to gravity, g = 10 m/s2 and Height, h = 850 m
Therefore, P.E. = m × g × h = 75 × 10 × 850 = 637500 J
(c) Total mechanical energy of the missile = K.E. + P.E.
= 135000 + 637500 = 7.7 × 105J
(d) K.E. at the ground = 7.7 × 105J
K.E. = 12mv2, Therefore, v2 = 2 K.E. / m
⇒ v = √2K.E.
m = √2×7.7×1075 5
m/s
Velocity with which missile strikes the ground =143.2 m/s
Question 2:
(a) Two bodies of equal masses move with uniform velocities v and 3v respectively. Find the ratio of their kinetic energies.
(b) An electric heater is rated 1500 W. How much energy does it use in 10 h? Express your answer in (i) kWh (ii) Joules [CBSE 2011]
Answer 2:
(a) Kinetic energy ∝ v2.
Therefore, for energies KE1 and KE2,
KE1 KE2 www.tiwariacademy.com A Free web support in education
= vv1222 = (3v)(v)2
2 = 9vv2 2
⟹ KEKE1
2 = 19
(b) Given,
Power, P = 1500 W = 1.5 kW, Time, t = 10 h
(i) Energy = Power × time = 1.5 × 10 = 15 kWh
(ii) 15 kWh = 15 × 3.6 × 106 J [ 1 kWh = 3.6 × 106 J ]
Question 3:
(a) Give reason for the following:
(i) The kinetic energy of a freely falling object increases, yet it holds law of conservation of energy.
(ii) A girl fills up 10 pages of a notebook in order to practice sums, yet she has not done ‘work’ in terms of Science / Scientific concept.
(iii) Work done by gravitational force on an object moved along a horizontal path, is zero.
(b) Find the energy in kWh consumed in 24 hours by two electric devices, one of 100 W and other of 500 W.
[CBSE 2010]
Answer 3:
(a)(i) According to the law of conservation of energy, energy can neither be created nor be destroyed but can be transformed from one form to another. Here, the increase in kinetic energy is due to transformation of potential energy into kinetic energy i.e.
Loss in P.E. = gain in K.E. So, the conservation of energy holds good.
(ii) There has been no displacement of the pages of the book so work done is zero. This is because W = F × s and when s = 0 then W = 0.
(iii) As the angle between the gravitational force and the displacement is 900. Hence, W = 0.
(b) Energy consumed= Power × time
Given,
Power of first device, P1 = 100W and Power of second device, P2 = 500W
Time taken for both the devices, t = 24 h
Total energy consumed= P1t + P2t = [100 × 24 + 500 × 24]
= [100 + 500] × 24 = 600 × 24 = 14400 Wh = 14.4 kWh
Question 4:
(a) Define kinetic energy of an object. Can kinetic energy of an object be negative? Give reason.
(b) A car weighing 1200 kg is uniformly accelerated from rest and covers a distance of 40 m in 5 s. Calculate the work, the car engine had to do during this time. [CBSE 2010]
Answer 4:
(a) Kinetic energy of an object is defined as the energy possessed by a body by virtue of its motion.
It No, is given the kinetic by, KE energy = 12mvof 2
an object cannot be negative since both mass of the object, m and velocity of a substance, v cannot be negative.
(b) Given,
Mass of the car, m = 1200 kg , Distance covered, s = 40 m
Time taken, t = 5 s, Initial velocity, v = 0
Work done, W =?
Since, W = F × s = ma × s [ Since, F = ma ]
For calculating work done, acceleration (a) is required.
Therefore 40 = 0 × t using, + 12a × 52
s = ut + 12at2
⟹ a = 40×2
25 = 3.2 m/s2
Now, W = F × s = ma × s
= 1200 × 3.2 × 40 = 153600 J
The work needed to be done by the car engine = 153600 J
Question 5:
(a) A battery lights a bulb. Describe the energy changes involved in the process.
(b) Calculate the amount of work needed to stop a car of 500 kg, moving at a speed of 36 km/h.
[CBSE 2010]
Answer 5:
(a) Energy conversion:
Chemical energy of battery → electrical energy → heat and light energy.
(b) Given,
Mass of a car, m = 500 kg,
Speed of car, v = 36 km/h = 10 m/s
Work needed to stop a car = K.E. of the car
Therefore,
K.E.= 12mv2 = 12 × 500 × 102
= 25000 Joule
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