2. Work and Energy
Complete solutions to the Exercises. Maharashtra State Board (Balbharati), Standard Nine Science and Technology.
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Exercises
(a) Explain the difference between potential energy and kinetic energy.
- Meaning: potential energy is the energy stored in an object because of its specific state or position, whereas kinetic energy is the energy an object has because of its motion.
- Formula: potential energy of an object at height h is P.E. = mgh, whereas kinetic energy of an object moving with velocity v is K.E. = ½mv².
- Depends on: potential energy depends on the mass, the height (position) or the state of the object; kinetic energy depends on the mass and the velocity of the object.
- Object at rest: an object at rest can have potential energy, but its kinetic energy is zero.
- Relative nature: potential energy depends on the reference level from which height is measured (for Ajay and Atul the same ball has different P.E.); kinetic energy depends on the speed of the object.
- Examples: water in an overhead tank, a stretched bow, a compressed spring and a wound-up clock spring have potential energy. A moving car, a flying arrow, a rolling ball and flowing water have kinetic energy.
(b) Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Suppose an object of mass m is initially at rest (u = 0). A force F acts on it and produces an acceleration a. After time t its velocity becomes v and its displacement is s.
By Newton’s second law: F = ma …… (1)
By the second equation of motion: s = ut + ½at²; with u = 0, s = ½at² …… (2)
∴ W = ma × ½at² = ½m(at)² …… (3) [from (1) and (2)]
By the first equation of motion: v = u + at; with u = 0, v = at
∴ v² = (at)² …… (4)
∴ W = ½mv² [from (3) and (4)]
The work done on the object is stored in it as kinetic energy.
∴ K.E. = ½mv²
(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.
Let an object of mass m be released from rest at point A, at height h above the ground. Let B be a point at distance x below A, and C the point on the ground.
Energies of a freely falling object at A, B and C. The total energy stays mgh throughout.
Total energy = 0 + mgh = mgh …… (1)
At B (fallen through x): u = 0, s = x, a = g
vB² = u² + 2as = 0 + 2gx = 2gx
K.E. = ½mvB² = ½m(2gx) = mgx; height = h − x, so P.E. = mg(h − x) = mgh − mgx
Total energy = mgx + mgh − mgx = mgh …… (2)
At C (on the ground): u = 0, s = h, a = g
vC² = 0 + 2gh = 2gh
K.E. = ½mvC² = ½m(2gh) = mgh; height = 0, so P.E. = 0
Total energy = mgh + 0 = mgh …… (3)
From (1), (2) and (3), the total energy is the same (mgh) at every point of the fall. As the object falls, its potential energy decreases and its kinetic energy increases by exactly the same amount. On reaching the ground the potential energy is zero and the kinetic energy is mgh — equal to the potential energy it had at the start. Hence the kinetic energy of a freely falling object on reaching the ground is nothing but its initial potential energy transformed into kinetic energy.
(d) Determine the amount of work done when an object is displaced at an angle of 30° with respect to the direction of the applied force.
Only the component F cos θ of the applied force acts along the displacement.
Let F be the applied force and s the displacement, with an angle θ = 30° between them. The component of the force in the direction of the displacement is F1 = F cos θ.
cos 30° = √3 / 2 ≈ 0.866
∴ W = (√3 / 2) F s ≈ 0.866 F s
So when the displacement is at 30° to the force, the work done is W = (√3/2) Fs ≈ 0.866 Fs, i.e. about 86.6% of the work that the same force would do if the displacement were along the force. For example, if F = 10 N and s = 2 m, W = 10 × 2 × 0.866 = 17.32 J.
(e) If an object has 0 momentum, does it have kinetic energy? Explain your answer.
No. An object with zero momentum has no kinetic energy.
- Momentum p = mv. The mass m of an object is never zero, so if p = 0, the velocity v must be zero — the object is at rest.
- Kinetic energy K.E. = ½mv². If v = 0, then K.E. = ½ × m × 0 = 0.
- The two quantities are directly related: K.E. = p² / 2m. When p = 0, K.E. = 0.
So an object with zero momentum is at rest and has zero kinetic energy. It may still have potential energy, depending on its position.
(f) Why is the work done on an object moving with uniform circular motion zero?
- In uniform circular motion, the force that keeps the object on the circle acts along the radius, towards the centre.
- At every point, the displacement of the object is along the tangent to the circle.
- The radius and the tangent are perpendicular to each other, so the angle between the force and the displacement is θ = 90°.
- W = Fs cos 90° = Fs × 0 = 0.
Therefore the work done on an object in uniform circular motion is zero. That is also why its speed, and hence its kinetic energy, does not change. For example, the gravitational force of the earth does no work on an artificial satellite moving in a circular orbit.
(a) For work to be performed, energy must be …
- (i) transferred from one place to another and (iii) transformed from one type to another
When work is done, energy passes from the object doing the work to the object on which work is done, and it usually changes form — for example, the chemical energy in our muscles becomes the potential energy of a bag we lift. Energy can never be destroyed, and simply concentrating it does no work.
(b) Joule is the unit of …
- (ii) work and (iv) energy
Energy is the capacity to do work, so both have the same unit, the joule. The unit of force is the newton and that of power is the watt.
(c) Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
- (ii) gravitational force and (iii) reaction force in vertical direction
The object does not move up or down, so the downward gravitational force (its weight) is balanced by the upward reaction of the surface. On a smooth surface there is practically no friction, and the horizontal applied force is not balanced by any force — it accelerates the object.
(d) Power is a measure of the …
- (i) the rapidity with which work is done
Power is the rate of doing work: P = W / t.
(e) While dragging or lifting an object, negative work is done by …
- (ii) gravitational force and (iii) frictional force
While lifting, the object moves up but gravity acts downward (θ = 180°), so gravity does negative work. While dragging, the object moves forward but friction acts backward, so friction does negative work. The applied force does positive work, and the vertical reaction force does zero work because it is perpendicular to the displacement.
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(a) The potential energy of your body is least when you are sleeping on the ground.
Potential energy depends on height (P.E. = mgh). When you lie on the ground, your body is at the lowest possible height, so its potential energy is the least.
(b) The total energy of an object falling freely towards the ground remains unchanged.
During free fall, potential energy is converted into kinetic energy, but their sum stays equal to mgh at every point (law of conservation of energy).
(c) If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy will not change.
Potential energy depends on height, not on speed. On a flat surface the height does not change, so the potential energy stays the same. (Its kinetic energy, however, becomes 4² = 16 times.)
(d) The work done on an object does not depend on the initial velocity of the object.
W = Fs cos θ depends only on the applied force, the displacement and the angle between them. The initial velocity does not appear in it.
(1) At the moment of releasing the balls, which energy do the balls have?
At the moment of release the balls are at rest at a height, so they have only potential energy (P.E. = mgh). Their kinetic energy is zero.
(2) As the balls roll down which energy is converted into which other form of energy?
As the balls roll down, their height decreases and their speed increases. So potential energy is converted into kinetic energy.
(3) Why do the balls cover the same distance on rolling down?
Both balls have the same mass and start from the same height, so they have the same potential energy at the top. The length of the channel does not matter. At the bottom, the whole of this potential energy has become kinetic energy, so both balls reach the floor with the same kinetic energy and the same speed. With equal energy they roll the same distance on the floor.
(4) What is the form of the eventual total energy of the balls?
At the bottom of the channels the total energy of the balls is in the form of kinetic energy. As they roll along the floor and finally stop, this kinetic energy is gradually changed into heat (and a little sound) because of friction.
(5) Which law related to energy does the above activity demonstrate? Explain.
The activity demonstrates the law of conservation of energy: energy can neither be created nor destroyed; it can only be converted from one form into another, and the total energy remains constant. The potential energy the balls have at the top is not lost; it is converted into an equal amount of kinetic energy at the bottom. That is why both balls, whatever the length of the channel, end up with the same energy and cover the same distance.
(a) An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?
Work done in one minute: W = P × t = 2000 × 60 = 120000 J
This work lifts the water: W = mgh
120000 = m × 9.8 × 10
m = 120000 ÷ 98 = 1224.5 kg
Answer: the pump lifts about 1224.5 kg of water every minute.
(b) If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?
Energy used per day = P × t = 1.2 × 0.5 = 0.6 kWh
Energy used in April = 0.6 × 30 = 18 kWh = 18 units
Answer: the iron consumes 18 units of electricity in April.
(c) If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound?
Energy left after 40% loss = 60% of mg × 10 = mg × 6
At the highest point of the rebound, all this energy is P.E. = mgh′
mgh′ = mg × 6 ⇒ h′ = 6 m
Answer: the ball will rebound to a height of 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?
v = 72 km/hr = 72 × 1000 ÷ 3600 = 20 m/s
m = 1500 kg
Work done = change in kinetic energy = ½mv² − ½mu² = ½m(v² − u²)
= ½ × 1500 × (20² − 15²)
= 750 × (400 − 225) = 750 × 175 = 131250 J
Answer: the work done on the car 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?
W = F × s = 10 × 0.3 = 3 J
Answer: Ravi did 3 J of work.
Some everyday examples to start with:
For each instance, note where the energy comes from, what form it changes into, and whether some of it is also wasted as heat or sound.