Std 9 Science Lesson 2 Work and Energy questions and answers
Exercises
1. Write detailed answers?
a. Explain the difference between potential energy and kinetic energy.
Ans.
Potential Energy | Kinetic Energy |
1. The energy stored in an object because of its specific state or position is called its potential energy. | 1. The energy which an object has because of its motion is called its kinetic energy. |
2. for e.g. an arrow is released from a stretched bow, water kept at a height flows through a pipe into the tap below, a compressed spring is released. | 2. for e.g. a fast cricket ball strikes the stumps, the striker hits a coin on the carom board, one marble strikes another in a game of marbles, etc |
3. P.E. = mgh | 3. K.E . = mv^{2} |
b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Ans. Suppose a stationary object of mass m moves because of an applied force.
Let u be its initial velocity (here u = 0).
Let the applied force be F.
This generates an acceleration a in the object, and, after time t, the velocity of the object becomes equal to v.
The displacement during this time is s.
The work done on the object,
W = F . s
W = F × s
According to Newton’s second law of motion,
F = ma ——– (1)
Similarly, using Newton’s second equation of motion
s = ut + 1/2 at^{2}
However, as initial velocity is zero, u = 0.
∴ s = 0 + 1/2 at^{2}
∴ s = 1/2 at^{2} ………..(2)
∴ W = ma × 1/2 at^{2} from eq. (1) and (2)
∴ W = 1/2 m × (at)^{2} ………..(3)
From Newton’s first equation of motion
v = u + at
∴ v = 0 + at
∴ v = at
∴ v^{2} = (at)^{2} ………..(4)
∴ W = 1/2 mv^{2} from equations (3) and (4)
The kinetic energy gained by an object is the amount of work done on the object.
∴ K.E = W
∴ K.E = 1/2 mv^{2}
c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Ans.
Let us look at the kinetic and potential energies of an object of mass m, falling freely from height h, when the object is at different heights,
As shown in the figure, the point A is at a height h from the ground.
Let the point B be at a distance x, vertically below A.
Let the point C be on the ground directly below A and B.
Let us calculate the energies of the object at A, B and C.
1. When the object is stationary at A,
its initial velocity is u = 0
∴ K.E = 1/2 mass × velocity^{2}
= 1/2 mu^{2}
K.E = 0
P.E = mgh
∴ Total energy = K.E + P.E
= 0 + mgh
Total Energy = mgh. – – – (1)
(2) Let the velocity of the object be v_{B} when it reaches point B, having fallen through a distance x.
u = 0, s = x, a = g
v^{2} = u^{2} + 2as
v_{B} ^{2} = 0 + 2gx
v_{B} ^{2} = 2gx
∴ K.E = 1/2 mv^{2}
= 1/2 m(2gx)
K.E = mgx
Height of the object when at B = h-x
∴ P.E = mg (h-x)
P.E = mgh – mgx
∴ Total Energy T.E. = K.E + P.E
= mgx + mgh – mgx
∴ T.E. = mgh ———- (2)
(3) Let the velocity of the object be v_{C} when it reaches the ground, near point C.
U = 0, s = h, a = g
v^{2} = u^{2} + 2as
v_{C} ^{2} = 0 + 2gh
∴ K.E = 1/2 mv_{C} ^{2}
= 1/2 m(2gh)
K.E = mgh
The height of the object from the ground at point C is h = 0
∴ P.E = mgh
= 0
∴ T.E. = K.E + P.E
T.E. = mgh —— (3)
From equations (1) (2) and (3) we see that the total potential energy of the object at its initial position is the same as the kinetic energy at the ground.
d. Determine the amount of work done when an object is displaced at an angle of 30^{0} with respect to the direction of the applied force.
W = Fs cos θ
∴ W = Fs cos 30 (∵ θ = 30°)
∴ W = Fs ……………. (∵ cos 30 = )
e. If an object has 0 momentum, does it have kinetic energy? Explain your answer.
Ans. No, the object does not have kinetic energy if it has 0 momentum.
Momentum = m x v
If momentum = 0, then v = 0, since mass cannot = 0
K.E = 1/2 mv^{2 }
Since v = 0, then K.E. = 0
Therefore, when momentum of an object is 0, the K.E. is 0.
f. Why is the work done on an object moving with uniform circular motion zero?
Ans. As the gravitational force acting on the object moving with uniform circular motion (along the radius of the circle) and its displacement (along the tangent to the circle) are perpendicular to each other, the work done by the gravitational force is zero.
i.e. θ = 90º and cos 90 = 0
∴ W = Fs cos θ
∴ W = 0
2. Choose one or more correct alternatives.
a. For work to be performed, energy must be ….
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed
Ans: transferred from one place to another.
b. Joule is the unit of …
(i) force
(ii) work
(iii) power
(iv) energy
Ans. Work and energy
c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction
Ans. gravitational force, reaction force in vertical direction
d. Power is a measure of the …….
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time
Ans. the rapidity with which work is done, the slowness with which work is performed
e. While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force
Ans. Gravitational force and frictional force
3. Rewrite the following sentences using proper alternative.
a. The potential energy of your body is least when you are …..
(i) sitting on a chair
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
Answer:
(iii) sleeping on the ground
b. The total energy of an object falling freely towards the ground …
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Answer:
(iii) increases
c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ….
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Answer:
(ii) will not change
d. The work done on an object does not depend on ….
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Answer:
(iii) initial velocity of the object
4. Study the following activity and answer the questions.
1. Take two aluminium channels of different lengths.
2. Place the lower ends of the channels on the floor and hold their upper ends at the same height.
3. Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance.
Questions
1. At the moment of releasing the balls, which energy do the balls have?
Ans. Potential energy.
2. As the balls roll down which energy is converted into which other form of energy?
Ans. Potential energy is converted to kinetic energy.
3. Why do the balls cover the same distance on rolling down?
Ans. Because they are moving at the same speed.
4. What is the form of the eventual total energy of the balls?
Ans. The total energy of the balls is Kinetic energy.
5. Which law related to energy does the above activity demonstrate? Explain.
Ans. The activity demonstrates the law of conservation of energy.
‘Energy can neither be created nor destroyed. It can be converted from one form into another. Thus, the total amount of energy in the universe remains constant.’
5. Solve the following examples.
a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Answer:
P = 2 kW
= 2 x 10^{3} watt,
t = 1 m = 60 seconds
h=10m.
g=9.8m/s^{2}.
mass of water(m) = ?
Solution:
P = W / t
P = mgh / t
2 x 10^{3}= m x 9.8 x 10 / 60
m = 2 x 10^{3} x 60 / 9.8 x 10
= 2 x 1000 x 60 / 98
= 120000 / 98
= 1224.48 = 1224.5 Kg
Ans. The amount of water lifted by the pump is 1224.5 Kg.