Chapter 2 – Work and Energy
1. Write detailed answers?
a. Explain the difference between potential energy and kinetic energy.
Ans:
| Potential energy | Kinetic energy |
|---|---|
(1) Potential energy is the energy possessed by the body because of its shape or position. | (1) Kinetic energy is the energy possessed by the body due to its motion. |
(2) Potential energy is given by the formula:
P.E = mgh. | (2) Kinetic energy is given by the formula:
K.E = \(\large \frac {1}{2}\) mv². |
(3) For example: Water at the top of a waterfall, before the drop. | (3) For example: Flowing water, such as when falling from a waterfall. |
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 = Fs …(i)
According to Newton’s second law of motion,
F = ma …(ii)
Similarly, using Newton’s second equation of motion
s = ut + \(\large \frac {1}{2}\) at²
However, as initial velocity is zero, u = 0,
∴ s = ut + \(\large \frac {1}{2}\) at²
∴ s = 0 + \(\large \frac {1}{2}\) at²
∴ s = \(\large \frac {1}{2}\) at² …(iii)
W = ma × \(\large \frac {1}{2}\) at² …[From (i) and (ii)]
∴ W = \(\large \frac {1}{2}\) m × (at)² …(iv)
Using Newton’s first equation of motion
v = u + at
∴ v = 0 + at
∴ v = at
∴ v² = a²t²
∴ v² = (at)² …(v)
∴ W = \(\large \frac {1}{2}\) m × v² …[From (iv) and (v)]
∴ W = \(\large \frac {1}{2}\) mv² …(vi)
The kinetic energy gained by an object is the amount of work done on the object.
∴ K.E = W
∴ K.E = \(\large \frac {1}{2}\) mv² …[From (vi)]
This is the formula for the kinetic energy of an object of mass ‘m’, moving with velocity ‘v’.
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, 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 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.
We know that,
K.E = \(\large \frac {1}{2}\) mv²
∴ K.E = \(\large \frac {1}{2}\) m(u)²
∴ K.E = 0
and P.E = mgh
Total energy = K.E + P.E
∴ Total energy = 0 + mgh
∴ Total Energy = mgh …(i)
(2) Let the velocity of the object be v\(_C\) when it reaches point B, having fallen through a distance x.
u = 0
s = x
a = g
We know that,
v² = u² + 2as
∴ v\(_B\)² = 0 + 2gx
∴ v\(_B\)² = 0 + 2gx
∴ v\(_B\)² = 2gx
∴ K.E = \(\large \frac {1}{2}\) mv²
∴ K.E = \(\large \frac {1}{2}\) m(2gx)
∴ K.E = mgx
Height of the object when at B = h – x
∴ P.E = mg (h – x)
∴ P.E = mgh – mgx
Now,
T.E. = K.E + P.E
∴ T.E. = mgx + mgh – mgx
∴ T.E. = mgh …(ii)
(3) Let the velocity of the object be vC when it reaches the ground, near point C.
U = 0
s = h
a = g
v² = u² + 2as
v\(_C\)² = 0 + 2gh
We know that,
K.E = \(\large \frac {1}{2}\) mv²
∴ K.E = \(\large \frac {1}{2}\) mv\(_C\)²
∴ K.E = \(\large \frac {1}{2}\) m(2gh)
∴ K.E = mgh
The height of the object from the ground at point C is h = 0
∴ P.E = mgh = 0
Now,
T.E. = K.E + P.E
T.E. = mgh + 0
T.E. = mgh …(iii)
From equations (i), (ii) and (iii) 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 300 with respect to the direction of the applied force.
Solution:
When object is displaced by displacement ‘S’ and by applying force ‘F’ at an ‘angle’ 30⁰
Work done can be given as,
W = Fs cos θ
∴ W = Fs cos 30° …(∵ θ = 30°)
∴ W = Fs (\(\large \frac {sqrt{3}}{2}\)) …(∵ cos 30° = \(\large \frac {sqrt{3}}{2}\))
Ans: The amount of work done when an object is displaced at an angle of 30° is Fs (\(\large \frac {sqrt{3}}{2}\))
e. If an object has 0 momentum, does it have kinetic energy? Explain your answer.
Ans: No, it does not have kinetic energy if it does not have momentum.
Momentum is the product of mass and velocity. If it is zero, it implies that v = 0 (since mass can never be zero).
Now,
K.E = \(\large \frac {1}{2}\) mv²
So if v = 0 then K.E also will be zero.
Thus, if an object has no momentum then it cannot possess kinetic energy.
f. Why is the work done on an object moving with uniform circular motion zero?
Ans:
(i) When a body performs uniform circular motion, then the force acting on it is along the radius of the circle.
(ii) While its displacement is along the tangent to the circle. Thus, they are perpendicular to each other.
Hence θ = 90º and cos 90 = 0
∴ W = Fs cos θ
∴ W = 0
Thus, the work done on an object moving with uniform circular motion is zero.
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: Options:
(i) transferred from one place to another
(iii) transformed from one type to another
b. Joule is the unit of ________
(i) force
(ii) work
(iii) power
(iv) energy
Ans: Options:
(ii) work
(iv) 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: Options:
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction
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: Options:
(i) the rapidity with which work is done
(iii) 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: Options:
(ii) gravitational force
(iii) frictional force
3. Rewrite the following sentences using proper alternatives.
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
Ans: Option (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
Ans: Option (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.
Ans: Option (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.
Ans: Option (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: At the moment of releasing the ball they possess P.E as they are at a height above the ground.
2. As the balls roll down which energy is converted into which other form of energy?
Ans: As the balls roll down, the P.E is converted into K.E since they are now in motion.
3. Why do the balls cover the same distance on rolling down?
Ans: Since they have been released from the same height, they will cover the same distance.
4. What is the form of the eventual total energy of the balls?
Ans: The eventual form of the total energy of the balls is “Mechanical Energy” i.e a combination of P.E and K.E.
5. Which law related to energy does the above activity demonstrate? Explain.
Ans: The above activity demonstrates the “Law of Conservation of Energy”.
As the balls roll down, the value of ‘h’ decreases due to which P.E also decreases as P.E = mgh. At the same time, the velocity of the balls increases as they roll down due to which their K.E increases as K.E = \(\large \frac {1}{2}\) mv².
Thus, we find that P.E decreases while K.E increases. But energy can neither be created nor destroyed. Thus, the P.E that decreases is in fact being converted to K.E.
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?
Given :
Power (P) = 2 kW = 2000 W
Height (h) = 10 m
Time (t) = 1 min = 60 s
Acceleration due to gravity (g) = 9.8 m/s2
To find :
Mass of water (m)= ?
Solution :
We know that,
P = \(\large \frac {mgh}{t}\)
∴ m = \(\large \frac {Pt}{gh}\)
∴ m = \(\large \frac {2000 \,×\, 60}{9.8 \,×\, 10}\)
∴ m = \(\large \frac {120000}{98}\)
∴ m = 1224.5 kg
Ans: Water lifted by the pump is 1224.5 kg
b. If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?
Given :
Power of electric iron (P) = 1200 W =
\(\large \frac {1200}{1000}\) kW = 1.2 kW
Time (t) = 30 min × 30 days
∴ Time (t) = 0.5 hr × 30 days
∴ Time (t) = 15 hr
To find :
Energy consumed = ?
Solution :
We know that,
Electric power = \(\large \frac {Electricity \,consumed}{time}\)
∴ Electricity consumed = Electric power × time
∴ Electricity consumed = 1.2 × 15
∴ Electricity consumed = 18 kWh
∴ Electricity consumed = 18 units
Ans: The units of energy consumed in the month of April by the iron is 18 units.
c. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound?
Given :
Initial height (h\(_1\)) = 10 m
Initial P.E = P.E\(_1\)
Final P.E = P.E\(_2\)
P.E\(_2\) = 60% of P.E\(_1\)
∴ P.E\(_2\) = 0.6 P.E\(_1\)
To Find :
Final height (h\(_2\)) = ?
Solution :
We know that,
P.E = mgh
∴ P.E\(_1\) = mgh\(_1\) …(i)
and P.E\(_2\) = mgh\(_2\) …(ii)
Dividing (ii) by (i), we get,
\(\large \frac {P.E_2}{P.E_1}\) = \(\large \frac {mgh_2}{mgh_1}\)
∴ \(\large \frac {0.6P.E_1}{P.E_1}\) = \(\large \frac {h_2}{10}\)
∴ 0.6 = \(\large \frac {h_2}{10}\)
∴ h\(_2\) = 0.6 × 10
∴ h\(_2\) = 6m
Ans: The ball will rebound by 6 m.
d. The velocity of a car increases from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg ?
Given :
Mass (m) = 1500 kg
Initial velocity (u) = 54 km/hr
(km/hr has to be converted to m/s)
∴ Initial velocity (u) = 54 × \(\large \frac {1000}{3600}\) m/s
∴ Initial velocity (u) = 54 × \(\large \frac {5}{18}\) m/s
∴ Initial velocity (u) = 15 m/s
Final velocity (v) = 72 km/hr
∴ Final velocity (v) = 72 × \(\large \frac {1000}{3600}\) m/s
∴ Final velocity (v) = 72 × \(\large \frac {5}{18}\) m/s
∴ Final velocity (v) = 20 m/s
To Find :
Work done to increase the velocity i.e. Change in K.E = ?
Solution :
Change in K.E = \(\large \frac {1}{2}\) mv² – \(\large \frac {1}{2}\) mu²
∴ Change in K.E = \(\large \frac {1}{2}\) m (v² – u²)
∴ Change in K.E = \(\large \frac {1}{2}\) × 1500 (20² – 15²)
∴ Change in K.E = 750 × (400 – 225)
∴ Change in K.E = 750 × 175
∴ Change in K.E = 131250 J
Ans: Work done to increase the velocity is 131250 J.
e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi?
Given :
Force (F) = 10 N
θ = 0º …(Since force and displacement are in same direction)
Displacement (s) = 30 cm = 0.3 m
To Find :
Work (W) = ?
Solution :
We know that,
W = Fs cos θ
∴ W = 10 × 0.3 × cos 0
∴ W = 3 × 1 …(∵ cos 0 = 1)
∴ W = 3 J
Ans: The work done by Ravi is 3 J.