NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work and Energy (MCQ, SAQ and LAQ)
Chapter Name | NCERT Exemplar Solutions for Class 9 Science Ch 11 Work and Energy |
Topics Covered |
|
Related Study |
|
Objective Type Questions for Work and Energy
1. When a body falls freely towards the Earth, then its total energy :
(a) Increases
(b) Decreases
(c) Remains constant
(d) First increases and then decreases
Solution
(c) Remains constant
2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car :
(a) Does not change
(b) Becomes twice to that of initial
(c) Becomes 4 times that of initial
(d) Becomes 16 times that of initial
Solution
(a) Does not change
3. In case of negative work the angle between the force and displacement is :
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Solution
(d) 180°
4. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same :
(a) Acceleration
(b) Momentum
(c) Potential energy
(d) Kinetic energy
Solution
(a) Acceleration
5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be : (g = 10 m s–2)
(a) 6 × 103 J
(b) 6 J
(c) 0.6 J
(d) zero
Solution
(d) zero
6. Which one of the following is not the unit of energy?
(a) Joule
(b) Newton metre
(c) Kilowatt
(d) Kilowatt hour
Solution
(c) Kilowatt
7. The work done on an object does not depend upon the :
(a) Displacement
(b) Force applied
(c) Angle between force and displacement
(d) Initial velocity of the object
Solution
(d) Initial velocity of the object
8. Water stored in a dam possesses :
(a) No energy
(b) Electrical energy
(c) Kinetic energy
(d) Potential energy
Solution
(d) Potential energy
9. A body is falling from a height h. After it has fallen a height 2 m, it will possess :
(a) Only potential energy
(b) Only kinetic energy
(c) Half potential and half kinetic energy
(d) More kinetic and less potential energy
Solution
(c) Half potential and half kinetic energy
Short Answer Questions for Work and Energy
10. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Solution
11. Avinash can run with a speed of 8 ms–1 against the frictional force of 10 N, and Kapil can move with a speed of 3 ms–1 against the frictional force of 25 N. Who is more powerful and why?
Solution
As work equals to product of force and distance, you can write the equation for power in the following way, assuming that the force acts along the direction of travel :
Power = Work done/Time
But, work done = Force × distance
Therefore, P = W/t = Fs/t
Where s is the distance travelled.
As Speed = Distance/Time
Therefore, the object’s speed, v is
P = W/t = Fs/t = Fv
Power of Avinash = 10 × 8 = 80 W
Power of kapil = 25 × 3 = 75 W
Therefore, Avinash is more powerful than Kapil.
12. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a roundabout figure of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.
Solution
F = 5 N
As work equals force times distance
W = f.s
W = 5 × [1500 + 200 + 2000]
= 18500 J
Here, 1.5 km = 1500 m; 2 km = 2000 m
13. Can any object have mechanical energy even if its momentum is zero ? Explain.
Solution
Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy. Mechanical energy = Potential energy+ Kinetic energy
Momentum = my
Given, Momentum is zero which means velocity is zero.
Therefore,
Kinetic energy = ½ mv2
As v = 0 (given)
Kinetic energy = ½ × m × 0
KE = 0
Mechanical energy = Potential energy + Kinetic energy
Mechanical energy = Potential energy + 0
Mechanical energy = Potential energy
Therefore, we can say that an object can have mechanical energy even if its momentum is zero.
14. The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given, g = 10 m s–2)
Solution
Power of the pump = 2kW = 2000 W;
t = 1 min = 60 sec;
height = 10 m;
g = 10 ms-2
Power = work done per unit time
(The force on an object of mass m at the surface of the earth is mg, from F = ma, when acceleration is g, the acceleration at the surface of the Earth. If the object falls through a distance h, then the work done on the object by the force of gravity is mg times h, force times distance.)
Work = mgh
P = mgh/t
2000 W = m × 10 ms-1 × 10 m
2000 W = ( m × 100 m)/60 s
m = (2000 × 60)/100
m = 1200 kg
Therefore, the pump can raise 1200 kg of water in one minute.
15. The weight of a person on a planet A is about half that on the Earth. He can jump up to 0.4 m height on the surface of the Earth. How high he can jump on the planet A?
Solution
As per the definition of force, the equation of force due to gravity is given by, W = mg, i.e., weight is equal to mass times gravitational acceleration In this case, the force is better known as the weight of the object. Weight of a person on Earth = w = mg1 (given);
height the person can jump (h1) = 0.4 m
U = potential energy from height,
m = mass of the object,
g = gravity and height = h
U = mgh
Therefore,
potential energy = mg1h1 ...(1)
where(h1) = 0.4 m; g1 = g
Now,
Weight of the person on planet A = W/2 = mg2/2
Let the height the person can jump = h2 :
g2 = ½ , g1 = ½ g ….(2)
Therefore,
Potential energy of planet A = mg2h2
Potential energy of the person on the Earth = mg1h1 …(3)
h2 = 0.4 × 2 = 0.8 m
Hence, he can jump double the height with the same muscular force.
16. The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
Solution
Consider that a force ‘F’ is applied on a body having mass ‘m’ and the distance travelled be ‘s’.
Work Done(joules) = Force (Newton) × Distance (meter)
W = F × s …(1)
As,
Force (F) = ma ...(2)
By substituting (2) in (1), we get
W = ma × s ….(3)
Using the Newton’s third equation of motion,
V2 – u2 = 2as 
W = final kinetic energy – initial kinetic energy
Work done = change in kinetic energy
17. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force ? Explain it with an example.
Solution
Yes, it is possible when force is acting perpendicular to the direction of displacement. For uniform circular motion, the force acts perpendicular to the direction of the motion and so, the force never does any work upon the object. E.g., an eraser tied to a string and moving the eraser in a circle at constant speed by holding onto the end of the string.
18. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2)
Solution
Given, g = 10 ms-2 ; h = 10 m
energy possessed by the ball = mgh
= m × 10 × 10 = 100 m joules
Energy left in the ball after striking the ground
= (100 – 40)/100 = 60/100
( As energy is reduced by 40% after striking the ground)
Therefore,
remaining energy = 60 m joules ….(1)
Let the height at which the ball bounces back = h1 ….(2)
Therefore,
Energy possessed by the ball = mgh
Using (1) and (2), we get
60 m = m × 10 × h1
h1 = 6 metres.
Thus, the height at which the ball bounces back = 6 m
19. If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April.
Solution
Given,
Power (P) = 1200 W = 1200/1000 = 1.2 KW:
Time (t) = 30 min = 30/60 hr = ½ hr = 0.5 hr
Number of days in april = 30 days
We need the time for a month (april) :
therefore, 0.5 × 30 = 15 h
Power = Energy/Time
Therefore, = Energy = Power × Time
E = 1.2 × 15 = 18 kWh
Long Answer Questions for Work and Energy
20 A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?Solution
Consider the mass and velocity of the light object to be m1 and v1 respectively. Similarly, let the mass and velocity of the heavy object to be m2 and v2 respectively
Momentum = mass × velocity
i.e., p = mv ...(1)
Momentum of light object = m1v1
Momentum of heavy object = m2v2
Given, light and a heavy object have the same momentum
Therefore, m1v1 = m2v2
But, we know that
Kinetic energy = ½ mv2
Thus,
Kinetic energy of light object (KE1) 
But, V1 > V2 [from (4)]
Therefore, (K.E.)l > (K.E.)h
‘The lighter one will have more kinetic energy than the heavy one. Moreover, Kinetic energy is directly proportional to the mass of the object.
21. An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now, car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.
Solution
Given, m(A) = m(B) = 1000 kg; Final velocity of ear
A(vA) = 0 (since it comes to rest after colliding with car B); Initial velocity of the car (uA) = 36 km/h = 10 m/s (since, 1 km/hr = 518 m/s);
Frictional force = 100 N
Since, the car A moves with a uniform speed, it means that the engine of car applies a force equal to the frictional force
Power = (Force × Distance)/Time
Distance = (Speed × Time)
Power = [Force × (Speed × Time)]/Time
Therefore, Power = Force × Speed
Power of car A = F v = 100 N × 10 m/s = 1000 W
Initial velocity of car B(ua) = 0;
Final velocity of car A(uA) = 0
By using the law of conservation of momentum,
mAuA + maua = mAvA + mava
Momentum before collision = momentum after collision
1000 × 10 + 1000 × 0 = 1000 × 0 + 1000 ×va = 10 ms-1
⇒ va = 10 ms-1
The speed of the car (B) just after the collision = 10 ms-1 .
22. A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl?
Solution
Given,
Mass of the girl = 35 kg;
Mass of the trolley = 5 kg;
Initial velocity(u) = 4 ms-1 ;
Final velocity (v) = 0 (as it comes to rest) ;
distance (s) = 16 m
Therefore, by using equation of motion, we get
v2 = u2 + 2as
⇒ 0 = (4)2 + 2 × a × (16)
⇒ 0 = 16 + 32a
⇒ −16 = 32a
⇒ −16/32 = a
⇒ −1/2 = a
⇒ − 0.5 m/s2 = a
(Acceleration is negative, therefore, retardation)
Force exerted (F) = ma
F = 40 × −1/2
⇒ F = −20N
(a) Work done = force × distance
W = F × d
⇒ W = mass (As F = ma)
Total mass of the trolley = mass of girl + mass of trolley
= 36 + 5 = 40 kg
Therefore,
Work done on the trolley = 40 × ½ × 16
= 20 × 16 = 320 J
(b) Work done by the girl = F × d × mass
Work done by the girl = 35 × 0.5 × 16
⇒ Work done by the girl = 280 J
23. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.
(a) How much work is done by the men in lifting the box?
(b) How much work do they do in just holding it?
(c) Why do they get tired while holding it? (g = 10 ms–2)
Solution
(a) Given, mass = 250 kg;
height(s) = 1 m;
g = 10 ms-2
F = mg(g = gravity)
⇒ F = 250 kg × g (g = 10 ms-2) = 2500 N
⇒ s = 1 m
⇒ W = F.s = 2500 × 1 N m = 2500 J
(b) The men did not do any work in just holding it because the box does not move at all. Therefore, kinetic energy is zero (when the box is held up, it is at rest, no movement, thus it has zero speed).
(c) In order to hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. They are working against gravity, air friction, etc. to hold the box up thereby making them feel tired.
24. What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 ms–2)
Solution
(i) Power is defined as the rate of doing work. Power measures the rate of work done, i.e., how fast or how slow the work is done.
Power = work/time
kW defines how much energy a device uses or generates in a given amount of time. Meanwhile kWh defines how much energy that device actually uses or generates.
kW is a measure of power while, kWh is a measure of energy.
kilowatt hour = kilowatt × hour
or kWh = kW × hr
Similarly, kilowatt = kilowatt hour/hour
or kW = kWh/ hr
(ii) Given, hr = 20 m, and
mass = 2000 × 103 kg = 2 × 106 kg
Power = (work done)/time taken
(here, work done = potential energy gained)
Power = mgh/t 
25. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 m s–1 vertically? g = 10 m/s–2 .
Solution
Given, Power = 100 W,
speed = 1 ms-1;
g = 10 m/s2
Power = Work done/Time taken
(here, work done = potential energy gained)
Therefore, Power = mgh/t ….(1)
But, h/t = speed at which the body is being lifted …(2)
By substituting (2) in (1), we get
Power = m × g × speed
Thus, m = power/(g × speed)
m = 100/(10×1)
m = 10 kg
26. Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s–1?
Solution
One watt is defined as the energy consumption rate of one joule per second. The power is said to be one watt, when a work of 1 joule is done in 1 s
1 W =1 J/ 1 s
One watt is also defined as the current flow of one ampere with voltage of one volt.
kilowatt = 1000 Js-1
Given, m = 150 kg;
power = 500 w/kg:
speed = 20 m/s
Power is 500 W/kg. So, total power developed by the engine of 150 kg.
Total power = 150 × 500 = 7.5 × 104 W
As Power = force × speed
Force = Power/Speed = (7.5 × 104)/20
= 3.75 × 103 N
27. Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 ms–2)
(i) A butterfly of mass 1.0 g that flies upward at a rate of 0.5 ms–1.
(ii) A 250 g squirrel climbing up on a tree at a rate of 0.5 ms–1.
Solution
(i) Given, mass of butterfly = 1 g = (1/1000) kg :
g = 10 ms-2 ;
speed (v) = 0.5 ms-1
Power = force × speed
But, Force = mg
Therefore, Power = mg × v
P = 1/1000 × 10 × 0.5
P = 0.5/100 = 5 × 103 W
(ii) Given, mass of squirrel = 250 g = 250/1000 g
= ¼ kg
g = 10 ms-2 ;
speed(v) = 0.5 ms-1
Power = force × speed
But, Force = mg
Therefore, Power = mg × v
P = ¼ × 10 × 0.5
P = 1.25 W
Thus, the power with which the squirrel is climbing is more than that of a butterfly flying.