Chapter - 4 Work and Energy

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Q
Work and Energy

Question:

Certain force acting on a 20 kg mass changes its velocity from 5 ms^{-1}to 2 ms^{-1}. Calculate the work done by the force.

Answer:

Given: Initial velocity, u = 5 ms^{-1}

Final velocity, v = 2 ms^{-1}

Mass of the body, m = 20\ kg

We know, kinetic energy of a body is

K.E. = frac{1}{2}mv^2

The initial kinetic energy of the body, 

E_i = frac{1}{2}mu^2

E_i = frac{1}{2} 	imes 20 	imes (5)^2

E_i = 250 J

The final kinetic energy of the body, 

E_f = frac{1}{2}mv^2

E_f = frac{1}{2} 	imes 20 	imes (2)^2

E_f = 40 J

Work done = Change in kinetic energy

Work done = Final K.E. - Initial K.E.

Work done = 40 J - 250 J

Work done = -210 J

Here, the negative sign shows that the force is acting in the opposite direction of the motion of the object.


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