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## Numerical Problems

#### 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. =   m v 2

W = mg       or        m =

m =   = 1200 kg

Thus               K.E. =   x 1200 x (20)2

= 600 X 400 = 240000 J

K.E. = 240 kJ

#### (56.25 J, 56.25 J)

Solution:     Mass of stone = m = 500 g =  kg = 0.5 kg

Velocity = v = 15 ms – 1

• Potential energy = P.E. = ?

(ii)               Kinetic energy = K.E. = ?

• Loss of K.E. = Gain in P.E.

m  –  m   = mgh

As the velocity of the stone at maximum height become zero, therefore, v= 0

x 0.5 x (0) –  x 0.5 x (15)2 = mgh

x 0.5 x 225 = mgh

– 56.25 = mgh

mgh = – 56. 25J

Since energy is always positive, therefore

P.E. = 56.25 J

(ii)                      K.E. =   m v 2

K.E. =  x 0.5 x (15)2

=  x 0.5 x 225

= 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. = ?

(i)                                           K.E. =   m v 2

K.E =   x 40 x (1.5)2

K.E =   x 40 x 2.25 = 45 J

(ii)                                          P.E = mgh

P.E = 40 x 10 x 6 = 2400 J

#### 6.6. A motor boat 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*S/t

P = 300*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*10*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*10*6/10

P = 1200 W

#### 6.10. An electric motor of 1hp is used to run water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and 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 = ?
• Power =

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 %