Q1: A force of 10N causes a displacement of 2 m in a body in its own direction. Calculate the work done by force.
Ans:
The work done by a constant force acting in the direction of displacement is given by
Work = F × s
= 10 N × 2 m
= 20 J
The work is positive because the force and displacement are in the same direction.
Q2: How much force is applied on the body when 150 joule of work is done in displacing the body through a distance of 10 m in the direction of force?
Ans:
Given W = 150 J and s = 10 m.
Using W = F × s, we get
F = W / s
= 150 J / 10 m
= 15 N
The force applied in the direction of displacement is 15 N.
Q3: A body of 5 kg raised to 2 m find the work done.
Ans:
The work done to raise a body against gravity equals the gain in gravitational potential energy (PE).
PE = mgh
= 5 kg × 9.8 m/s2 × 2 m
= 98 J
Therefore, the work done in raising the body is 98 J.
Q4: A work of 4900 J is done on load of mass 50 kg to lift it to a certain height. Calculate the height through which the load is lifted.
Ans:
Work done in lifting = increase in potential energy = mgh.
4900 J = 50 kg × 9.8 m/s2 × h
h = 4900 / (50 × 9.8) = 4900 / 490 = 10 m
The load is lifted through a height of 10 m.
Q5: An engine work 54,000 J work by exerting a force of 6000 N on it. What is the displacement of the force?
Ans:
Displacement s = W / F
= 54000 J / 6000 N= 9 m
Thus, the displacement is 9 m.
Q6: A body of mass 2 kg is moving with a speed of 20 m/s. Find the kinetic energy.
Ans:
Kinetic energy (KE) = ½ mv2
= 0.5 × 2 kg × (20 m/s)2
= 0.5 × 2 × 400
= 400 J
The kinetic energy of the body is 400 J.
Q7: A hammer of mass 1 kg falls freely from a height of 2 m. Calculate (I) The velocity and (II) The kinetic energy of the hammer just before it touches the ground. Does the velocity of hammer depend on the mass of hammer?
Ans:
Use energy conservation (neglecting air resistance): loss of potential energy = gain in kinetic energy.
PE at top = mgh = 1 kg × 9.8 m/s2 × 2 m = 19.6 J
Thus KE just before impact = 19.6 J.
Let v be the speed just before impact. Then ½ m v2 = 19.6 J.
19.6 = 0.5 × 1 × v2 → v2 = 39.2 → v ≈ 6.26 m/s.
(I) Velocity just before touching ground ≈ 6.26 m/s.
(II) Kinetic energy just before impact = 19.6 J.
The velocity does not depend on the mass of the hammer (neglecting air resistance) because v for a freely falling object from rest after falling height h is v = √(2gh), which contains no mass term.
Q8: A weight of 50 kg runs up a hill rising himself vertically 10 m in 20 sec. Calculate power. (Given g = 9.8 ms-1)
Ans: Work done = increase in potential energy = mgh
= 50 kg × 9.8 m/s2 × 10 m
= 4900 J.Power = Work / time = 4900 J / 20 s = 245 W.
The required power is 245 W.
Q9: A rickshaw puller pulls the rickshaw by applying a force of 100 N. If the rickshaw moves with constant velocity of 36 kmh-1. Find the power of rickshaw puller.
Ans:
Force = 100 N.
Velocity = 36 km/h = 36 × 5 / 18 = 10 m/s.
Power = Force × Velocity = 100 N × 10 m/s = 1000 W.
Therefore, the rickshaw puller supplies 1000 W of power. At constant velocity this power balances resistive forces such as friction and air resistance.
Q10: An athlete weighing 60 kg runs up a staircase having 10 steps each of 1 m in 30 sec. Calculate power (g = 9.8ms-1).
Ans:
Total vertical height climbed h = 10 × 1 m = 10 m.
Work done = increase in potential energy = mgh = 60 kg × 9.8 m/s2 × 10 m = 5880 J.
Power = Work / time = 5880 J / 30 s = 196 W.
The athlete’s power while climbing is 196 W.
Q11: A horse exert a force of 200N to pull the cart. If the horse cart system moves with velocity 36 kmh-1 on the level road, then find the power of horse in term of horse power (1 HP = 746 W).
Ans:
Velocity = 36 km/h = 10 m/s.
Power = F × v = 200 N × 10 m/s = 2000 W.
In horsepower: 2000 W ÷ 746 W per HP ≈ 2.68 HP.
So the horse develops about 2.68 horsepower.
Q12: An electric kettle of 500W is used to heat water everyday for 2 hours. Calculate the number of unit of electrical energy consumed by it in 10 days
Ans:
Power = 500 W = 0.5 kW. Daily usage = 2 h, for 10 days total time = 2 × 10 = 20 h.
Energy = P × t = 0.5 kW × 20 h = 10 kWh = 10 units.
The kettle consumes 10 units of electrical energy in 10 days.
Q13: Calculate the cost of using a 2 kWh immersion rod for heating water in a house for one hour each day for 60 days if the rate is 1.50 per unit kWh.
Ans:
Power = 2 kW, daily time = 1 h, period = 60 days → total energy = 2 kW × 60 h = 120 kWh = 120 units.
Cost = 120 units × Rs 1.50 per unit = Rs 180.
The total cost is Rs 180.
Q14: In an experiment to measure his power, a student records the time taken by him in running up a flight of steps on a staircase.
Ans:
Use the following data to calculate the power of the student :
Number of steps = 28,
Height of each step = 20 cm,
Time taken = 5.4 s,
Mass of student = 55 kg,
Acceleration due to gravity = 9.8 ms-2
The total vertical height climbed = 28 × 0.20 m = 5.6 m.
Work done = mgh = 55 kg × 9.8 m/s2 × 5.6 m = 3018.4 J.
Power = Work / time = 3018.4 J / 5.4 s ≈ 559 W.
The student’s power while climbing the steps is approximately 559 W.
Q15: The power of a heart which beats 72 times in a minute is 1.2 kW. Calculate the work done by heart for each beat. (1 kJ)
Ans:
Power P = 1.2 kW = 1200 W. Time interval considered = 1 minute = 60 s.
Total work done in 60 s = P × t = 1200 W × 60 s = 72000 J.
Number of beats in 60 s = 72. Work done per beat = 72000 J / 72 = 1000 J = 1 kJ.
Work done by the heart in each beat is 1 kJ.
Q16: Calculate the electricity bill amount for a month of 31 days, if the following devices are used as specified.
(a) 3 bulbs of 40 W for 6 hours.
(b) 4 tubelights of 50 W for 8 hours,
(c) A TV of 120 W for 6 hours.
The rate of electricity is Rs 2.50 per unit.
Ans:
Convert powers to kW and find daily energy consumption:
E1 (3 bulbs) = 3 × 40 W × 6 h = 720 Wh = 0.72 kWh per day.
E2 (4 tubelights) = 4 × 50 W × 8 h = 1600 Wh = 1.60 kWh per day.
E3 (TV) = 120 W × 6 h = 720 Wh = 0.72 kWh per day.
Total per day = 0.72 + 1.60 + 0.72 = 3.04 kWh per day.
For 31 days: Energy = 3.04 kWh × 31 = 94.24 kWh (units).
Cost = 94.24 × Rs 2.50 = Rs 235.60.
The electricity bill for the month is Rs 235.60.
Q17: (a) What is meant by mechanical energy? State its two forms. State the law of conservation of energy. Give an example in which we observe a continuous change of one form of energy into another and vice-versa.
(b) Calculate the amount of work required to stop a car of 1000 kg moving with a speed of 72 km/h.
Ans:
(a) Mechanical energy of a body is the sum of its kinetic energy and gravitational potential energy. Its two forms are:
– Kinetic energy (energy of motion).
– Potential energy (energy due to position in a force field, e.g., gravity).
Law of conservation of energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another.
Example: In a simple pendulum, energy continuously converts between kinetic and potential energy. At the mean position the energy is mainly kinetic, and at the extreme positions it is mainly potential. During oscillation the total mechanical energy (neglecting friction) remains constant.
(b) Given: m = 1000 kg, initial speed u = 72 km/h = 20 m/s, final speed v = 0.
Work required to stop = change in kinetic energy = ½ m (v2 – u2) = -½ m u2.
Magnitude of work done by the brakes = ½ × 1000 kg × (20 m/s)2 = 0.5 × 1000 × 400 = 200000 J = 2 × 105 J.
The brakes must do 2 × 105 J of work (energy removed) to bring the car to rest.