Question

A single force acts on a particle in rectilinear motion. A polt of velocity versus time for the motion is given. Then prove whether work done by the force on the particle in the interval is negative or positive?

Answer 1

The relation between work done and the kinetic energy is given as,

$\Delta K.E.=W$

In AB interval:

The change in the $K.E.$is $+ve$ so the work done will be $+ve$

In $BC$ interval:

Since the initial velocity is equal to the final velocity as shown in Fig.

Therefore, the workdone will be Zero as the change in $K.E.$ is zero.

In $CD$ interval:

The change is the $K.E.$is $-ve$ so the work done will also be $-ve$

In $DE$ interval:

The change in $K.E.$ is $+ve$ so the work done will also be $+ve$