Physics 9-Ch6 (Work and Energy)-Numerical Problems

 

Table of Contents

• A man has pulled a cart through 35 m applying a force of 300 N
• A block weighing 20 N is lifted 6 m vertically upward.
• Find its kinetic energy. (240 kJ)
• (i) P.E. at its maximum height  (ii) K.E. when it hits the ground
• Find the kinetic energy and the potential energy of the cyclist.
• A motor boat moves at a steady speed
• Find the power used by him to pull the block.
• A 50 kg man moved 25 steps up in 20 seconds. Find his power,
• Calculate the power of a pump that can lift
• An electric motor of 1hp is used to run the water pump.

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 = $�\times �$
                                    W = $300\times 35$
                                    W = 10500 J

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. = $(��)\times ℎ$
Thus,                         P.E. = $(2\times 10)\times 6$
                                  P.E. = 120 J

A car weighing 12 k N has a speed of 20 $�{�}^{-1}$. Find its kinetic energy. (240 kJ)

Solution:      

Weight of the car = w = 12kN = $12\times 1000$ N = 12000 N
Speed of the car = v= 20 $�{�}^{-1}$ 
Kinetic energy = K.E. = ?

K.E. = $\frac{1}{2}�{�}^2$
W = mg or m = $\frac{�}{�}$
m = $\frac{12000}{10}$ = 1200 kg

Thus, K.E. = $\frac{1}{2}\times 1200\times (20{)}^2$
= $600\times 400$ = 240000 J
K.E. = 240 kJ

A 500 g stone is thrown up with a velocity of 

$15�{�}^{-1}$. 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 = $\frac{500}{1000}$ kg = 0.5 kg

Velocity = v = 15 $�{�}^{-1}$ 

1. Potential energy = P.E. = ?
2. Kinetic energy = K.E. = ?
3. Loss of K.E. = Gain in P.E.

 

$\frac{1}{2}�{�}_{\mathrm{\_}}{�}^2$ – $\frac{1}{2}��\mathrm{\_}{�}^2$  = mgh

As the velocity of the stone at maximum height become zero, therefore, ${�}_{�}$  = 0

$\frac{1}{2}\times 0.5\times (0)-\frac{1}{2}\times 0.5\times (15{)}^2$ = mgh

$\frac{-1}{2}\times 0.5\times 225$ = mgh

$-$56.25 = mgh

mgh = $-$56.25J

 

Since energy is always positive, therefore

P.E. = 56.25 J

K.E. = $\frac{1}{2}�{�}^2$

K.E. = $\frac{1}{2}\times 0.5\times (15{)}^2$

= $\frac{1}{2}\times 0.5\times 225$

= 56.25 J

On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 $�{�}^{-1}$. 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 $�{�}^{-1}$ 
                  Mass of cyclist and the bicycle = m = 40 kg
            Kinetic energy = K.E. = ?
            Potential energy = P.E. = ?
            
(i) K.E. = $\frac{1}{2}�{�}^2$
    K.E = $\frac{1}{2}\times 40\times (1.5{)}^2$
    K.E = $\frac{1}{2}\times 40\times 2.25$ = 45 J
    
(ii) P.E = mgh
     P.E = $40\times 10\times 6$ = 2400 J

A motor boat moves at a steady speed of 4 $�{�}^{-1}$. Water-resistance acting on it is 4000 N. Calculate the power of its engine. (16 kW)

Solution:         

     Speed of the boat = v = $4�{�}^{-1}$ 
                         Force = F = 4000 N
                         Power = P = ?

                       P = F v
                       P = $4000\times 4$
                       P = 16000 W
                       P = $16\times 10^3$ W
                       P = 16 kW

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 =$\frac{����}{����}$ = $\frac{�}{�}$= $\frac{(�\times �)}{�}$
                         P =$\frac{(300\times 50)}{60}$ = $5\times 50$ = 250 W

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 = $25\times 16$ = 400 cm = $\frac{400}{100}$ m = 4 m
Time = t = 20 s
Power = P = ?

Power =$\frac{����}{����}$=$\frac{�}{�}$ = $\frac{��ℎ}{�}$
P =$\frac{(50\times 10\times 4)}{20}$
P = 100 W

Calculate the power of a pump that 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 =$\frac{����}{����}$ = $\frac{�}{�}$ = $\frac{��ℎ}{�}$
                     P = $\frac{(200\times 10\times 6)}{10}$ 
                     P = 1200 W

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 

$��{�}^{-3}$ )
(Mass of 1 liter of water = 1 kg) (447600 J, 26.8%)

Solution:            

Power = P = 1 hp = 746 W

                  Time = t = 10 min = $10\times 60$ s = 600 s

                  $\frac{��������}{������}$ = V = 800 liters

                  Height = h = 15 m

 

Work done = W = ?

Efficiency = E = ?

 

Power = $\frac{����}{����}$

             P =$\frac{�}{�}$

 

Or          W = $�\times �$

              W = $746\times 600$

              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 = $800\times 10\times 15$ = 120000J

 

                      % Efficiency    = $\frac{������}{�����}\times 100$

                  % Efficiency  =  $\frac{120000}{447600}\times 100$

                      Efficiency = 26. 8 %