Q 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
(c) remains constant
Q 2 – In case of negative work the angle between the force and displacement is
(a) 0°
(b) 45°
(c) 90°
(d) 180°
(d) 180°
Q 3 – 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
(a) does not change
Q 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) momenta
(c) potential energy
(d) kinetic energy
(a) acceleration
Q 5 – Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour
(b) newton metre
Q 6 – Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy
(d) potential energy
Q 7 – A body is falling from a height h. After it has fallen a height h/2, 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
(c) half potential and half kinetic energy
Q 8 – Work done by friction
(a) increases kinetic energy of body
(b) decreases kinetic energy of body
(c) increases potential energy of body
(d) decreases potential energy of body
(b) decreases kinetic energy of body
Q 9 – Fill in the blanks: [5]
- Work done is product of _______ and distance moved the direction of the force.
- If the angle between force and displacement is _____ then work done is said to be zero.
- Energy of a body is defined as the capacity or ______ to do work.
- Mechanical energy includes _______ and _________
- Energy can neither be _______ nor ______ it can only transformed from one form to another.
- force
- 90°
- ability
- kinetic energy, potential energy
- created, destroyed
Q 10 – Match the following: [8]
Column A | Column B |
(a) SI unit of power | (i) 1/2 mv2 |
(b) Kinetic energy | (ii) The change of one form of energy into another |
(c) Potential energy | (iii) watt |
(d) SI unit of work | (iv) If applied force on an object and displacement is in opposite direction. |
(e) Negative work | (v) Energy possessed by a body due to its position or configuration. |
(f) Power | (vi) Joule |
(g) Expression of kinetic energy | (vii) Energy possessed by a body due to its motion. |
(h) Transformation of energy | (viii) Rate of doing work |
Column A | Column B |
(a) SI unit of power | (iii) watt |
(b) Kinetic energy | (vii) Energy possessed by a body due to its motion. |
(c) Potential energy | (v) Energy possessed by a body due to its position or configuration. |
(d) SI unit of work | (vi) Joule |
(e) Negative work | (iv) If applied force on an object and displacement is in opposite direction. |
(f) Power | (viii) Rate of doing work |
(g) Expression of kinetic energy | (i) 1/2 mv2 |
(h) Transformation of energy | (ii) The change of one form of energy into another |
Q 11 – The kinetic energy of a car is 7000 J as it travels along a horizontal road. How much work is required to stop the car in 20 s?
The theorem states that the net work done by the force is equal to the change in kinetic energy of the object.
So, we can say :
w = ΔKE
Since the car finally comes to a stop, its final kinetic energy will be 0 Joules.
W = KE1 – KE2
W = 0 -7000 = -7000 joule
So, the work done to stop the car is -7000 Joule.
Q 12 – A coolie is walking on a railway platform with a load of 27 kg on his head. What is the amount of work done by him?
Work done by the coolie is zero, as W = Fs cos 90° = 0.
Q 13 – An electric heater of 1000 W is used for two hours in a day? What is the cost of using it for a month of 28 days, if one unit costs 3.00 rupees?
.Energy consumed by heater of 1000 W for 2 h in one
day =1000/1000 X 2 = 2 units
Number of units = Watt x hours x day / 1000
Energy consumed by heater in 28 days = 2 x 28 = 56 units.
Cost of 1 unit= Rs 3
Cost of 56 units = 3 x 56 = Rs 158
Q 14 – A boy stands on the edge of a cliff and throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it has 1000 J of kinetic energy. If he were to throw the stone horizontally outward from the cliff with the same initial speed of 10 m/s, how much kinetic energy would it have just before it hits the ground?
Kinetic Energy at the time of hitting the ground= Potential energy at the edge of cliff + Initial Kinetic energy
when the stone is thrown vertically downward
K 1 = m g h + 1 /2 m v2
= 1000 J ——-(1)
Now when the stone is thrown horizontally outward with the same speed
K 2 = m g h + 1/ 2 m v2
From equation
K 2 = 1000 J
Thus the K.E of the stone before it hits the ground is 1000 J
Q 15 – Calculate the work done required to be done to stop a car of 1500 kg motor at a velocity of 60 1/h.
When a object is moving with a constant velocity, it possess K.E
K.E = × m × v2
So, in order to bring the object to rest i.e its same magnitude of energy is required so, as the final energy comes to zero.
V = 60km/hr = 16.66m/s
So, magnitude of work that needs to be done = 1/2 × m × v2
Hence, work required to stop the car × 1500 × 16.662
Q 16 – What types of energy transformation takes place in the following:
(i) Electric heater
(ii) Solar battery
(iii) Dynamo
(iv) Steam engine and
(v) Hydroelectric power station?
Electric heater: Electric energy into heat energy.
Solar battery: Solar energy into electric energy.
Dynamo: Mechanical energy into electric energy.
Steam engine: Heat energy to mechanical energy.
Hydroelectric power station: Mechanical energy into electric energy.
Q 17 – Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified:
a. two bulbs of 40W for 6 hours.
b. two tubelights of 50W for 8 hours
c. A TV of 120W for 6 hours.
d. Give the rate of electricity is 2.50 rupees per unit?
Electrical energy consumed by two bulbs in 30 days = 2 × 40 × 6 × 30
= 14400 watt.hour
= 14.4 KWH
Electrical energy consumed by two tube lights in 30 days = 2 × 50 × 8 × 30
= 24000 watt.hr
= 24 KWH
Electrical energy consumed by a TV in 30 days = 120 × 6 × 30
= 21600 watt.hr
= 21.6 KWH
As we know, 1 KWH = 1 unit, and the cost per unit is given Rs. 2.50
The total electricity bill of one month = (14.4 + 24 + 21.6) × 2.50
= Rs. 150
Therefore, The total electricity bill of one month is Rs. 150
Q 18 – Briefly describing the gravitational potential energy, deduce an expression for the gravitational potential energy of a body of mass m placed at a height h, above the ground.
When an object is raised through a certain height above the ground, its energy increases. This is because the work is done on it, against gravity. The energy present in such an object is called gravitational potential energy. Thus, the gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity. Consider a body of mass m lying at point P on the Earth’s surface, where its potential energy is taken as zero. As weight, mg acts vertically downwards, so to lift the body to another position Q at a height h, we have to apply a minimum force which is equal to mg in the upward direction. Thus, work is done on the body against the force of gravity.
We know that,
Work done, W = Fs …(i)
As F = mg and s = h
Putting these values in equation (i), we get
W = mg × h = mgh
This work done on the body is equal to the gain in energy of the body. This is the potential energy of the body.
∴ Potential energy PE = mgh
Q 19 – 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)
(a) F = 250 kg × 10 ms-2 (g = 10 ms-2)
= 2500 N
s = 1 m
W = F.s
= 2500 N × 1 m
= 2500 Nm = 2500 J
(b) Zero, as the box does not move at all while holding it.
(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. While applying the force, muscular effort is involved. So, they feel tired.
Q 20 – An automobile engine propels a 1,000 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.
mA = mB = 1000 kg. v= 36 km/h =10 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 = F.v
= 100N × 10m/s = 1000W
After collision,
mAuA + mBuB = mAvA + mBvB
1000 × 10 + 1000 × 0 = 1000 × 0 + 1000 × vB
vB = 10 ms-1.