Class 9 Physics Chapter 6 numerical problems are according to the new syllabus of the Punjab Board. Chapter 6 is related to Work and Energy in which you will study the work, power, energy, types of energy, and efficiency of matter.
Class 9 Physics Chapter 6 Numerical Problems
The problems have been solved with easy-to-use methods using the following formulae:
W = F x S
P.E. = mgh
K.E. = 1/2 m v 2
P = F v
Numerical Problems Chapter 6 – Work and Energy
6.1. A man has pulled a cart through 35 m applying a force of 300 N. Find the work done by the man. (10500 J)
Solution:
Distance = S = 35 m
Force = F = 300 N
Work done = W = ?
W = F x S
W = 300 x 35
W = 10500 J
6.2. A block weighing 20 N is lifted 6 m vertically upward. Calculate the potential energy stored in it. (120 J)
Solution:
Weight of the block = W = 20 N
Height = h = 6 m
Potential energy = P. E. = ?
P.E. = mgh
We know that w = mg
P.E. = (mg) x h
Thus,
P.E. = (2 x 10) x 6
P.E. = 120 J
6.3. A car weighing 12 k N has a speed of 20 ms – 1. Find its kinetic energy. (240 kJ)
Solution:
Weight of the car = w = 12kN = 12 x 1000 N = 12000 N
Speed of the car = v= 20 ms – 1
Kinetic energy = K.E. = ?
K.E. = 1/2 m v 2
w = mg or m = w/g
m = 12000/10 = 1200 kg
Thus,
K.E. = 1/2 x 1200 x (20)2
K.E. = 600 X 400
K.E. = 240000 J
K.E. = 240 kJ
6.4. A 500 g stone is thrown up with a velocity of 15ms – 1. Find its
(i) P.E. at its maximum height
(ii) K.E. when it hits the ground (56.25 J, 56.25 J)
Solution:
Mass of stone = m = 500 g = 500/1000 kg = 0.5 kg
Velocity = v = 15 ms – 1
Potential energy = P.E. =?
Kinetic energy = K.E. =?
Loss of K.E. = Gain in P.E.
1/2mvf2 – 1/2mvi2 = mgh
As the velocity of the stone at maximum height become zero, therefore, vf = 0
1/2 x 0.5 x (0) – 1/2 x 0.5 x (15)2 = mgh
-1/2 x 0.5 x 225 = mgh
-56.25 = mgh
mgh = – 56. 25J
Since energy is always positive, therefore
P.E. = 56.25 J
K.E. = 1/2 m v 2
K.E. = 1/2 x 0.5 x (15)2
K.E. = 1/2 x 0.5 x 225
K.E. = 56.25 J
6.5. On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 ms – 1. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40 kg. (45 J, 2400 J)
Solution:
Height of the slope = h = 6 m
Speed of the cyclist = v = 1.5 ms – 1
Mass of cyclist and the bicycle = m = 40 kg
Kinetic energy = K.E. =?
Potential energy = P.E. =?
K.E. = 1/2 m v 2
K.E. = 1/2 x 40 x (1.5)2
K.E = 1/2 x 40 x 2.25 = 45 J
P.E = mgh
P.E = 40 x 10 x 6 = 2400 J
6.6. A motorboat moves at a steady speed of 4 ms – 1. Water resistance acting on it is 4000 N. Calculate the power of its engine. (16kW)
Solution:
Speed of the boat = v = 4ms – 1
Force = F = 4000 N
Power = P =?
P = F v
P = 4000 x 4
P = 16000 W
P = 16 x 103 W
P = 16 kW
6.7. A man pulls a block with a force of 300 N through 50 m in 60 s. Find the power used by him to pull the block. (250 W)
Solution:
Force = F = 300 N
Distance = S = 50m
Time = t = 60 s
Power = P =?
Power = work / time = W/t = F x S/t
P = 300 x 50 / 60 = 5 x 50 = 250 W
6.8 A 50 kg man moved 25 steps up in 20 seconds. Find his power, if each step is 16 cm high. (100 W)
Solution:
Mass = m = 50 kg
Total height = h = 25 x 16 = 400 cm = 400/100 m = 4 m
Time = t = 20 s
Power = P =?
Power = work/time = w/t = mgh/t
P = 50 x 10 x 4 / 20
P = 100 W
6.9. Calculate the power of a pump which can lift 200 kg of water through a height of 6 m in 10 seconds. (1200 watts)
Solution:
Mass = m = 200 kg
Height = h = 6m
Power = P =?
Power = work/time = w/t = mgh/t
P = 200 x 10 x 6 / 10
P = 1200 W
6.10. An electric motor of 1hp is used to run the water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and a height of 15 m. Find the actual work done by the electric motor to fill the tank. Also, find the efficiency of the system. (Density of water = 1000 kgm – 3 ) (Mass of 1 liter of water = 1 kg) (447600 J, 26.8 %)
Solution:
Power = P = 1 hp = 746 W
Time = t = 10 min = 10 x 60 s = 600 s
Capacity/volume = V = 800 liters
Height = h = 15 m
Work done = W =?
Efficiency = E =?
P = work/time
Or W = P x t
W = 746 x 600
W = 447600 J
Since the work done by the electric pump to fill the tank is 447600 J. It is equal to the input.
Hence input = actual work doe = W = 447600 J
Output = P.E = mgh
Since, 1 litre = 1kg, therefore 800 litres = 800 kg
Output = P.E = 800 x 10 x15 = 120000J
% Efficiency = output/input * 100
% Efficiency = 120000/447600 * 100
Efficiency = 26.8 %
Class 9 Physics Full Book Numerical Problems with Solution
