The velocity of a body moving in a straight

Question:

The velocity of a body moving in a straight line is increased by applying a constant force $F$, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Solution:

Consider a body or an object of mass $m$ moving with velocity $u$. Let a force $F$ be applied on the body so that the velocity attained by the body after travelling a distance $S$ is $v$ (Figure 5 ).

Work done by the force on the body is given by

$W=F S$ $\ldots$ (i)

Since velocity of the body changes so the body is accelerated. Let a be the acceleration of the body.

Therefore, according to Newton's second law of motion,

$F=m a$ ...(ii)

Using eqn. (ii) in eqn. (i), we get

$\mathrm{W}=(m a) \mathrm{S}$ $\ldots($ iii $)$

Now, using $v^{2}-u^{2}=2 a S$, we get

$\mathrm{S}=\frac{v^{2}-u^{2}}{2 a}$ $\ldots(i v)$

Using eqn. (iv) in eqn. (iii), we get

$\mathrm{W}=m a\left(\frac{v^{2}-u^{2}}{2 a}\right)=\frac{1}{2} m\left(v^{2}-u^{2}\right)$

or

$\mathrm{W}=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$ $\ldots(1)$

or

$W=$ Final K.E of body - Initial K.E of body

$=$ Change in K.E. of the body.

Thus, work done by a force on a body is equal to the change in kinetic energy of the body This is known as work-energy theorem.

 

Leave a comment