# Class 9 Physics Chapter 6 Numerical Problems 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

#### (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

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