Question

Force acting on a particle moving in a straight line varies with the velocities of the particle as $F=K.V$. Where $K$ is constant. The work done by this force in time $t$ is

• A. $KVt$
• B. $K^2{V}^2{t}^2$
• C. $K^{2}Vt$
• D. $K{V}^{2}t$

Answer 1

The correct option is D $K{V}^{2}t$

Given that,

Force $F=KV$

Velocity=$V$

We know that,

$F=Ma$

Now,

$Ma=KV$

$M\left(\frac{dV}{dt}\right)=KV$

$MdV=KVdt$

Multiply by v in both side

$MVdV=K{V}^{2}dt$

On integrating both side

$\underset{v_1}{\overset{v_2}{\int }}MVdV=\underset{0}{\overset{t}{\int }}K{V}^{2}dt$

$M[\frac{V_2^2}{2}-\frac{V_1^2}{2}]=K{V}^{2}t$

$\frac{M}{2}[V_2^2-V_1^2]=K{V}^{2}t$

Now, the change in K.E

Now, from work energy theorem

Change in energy = work done

$K.E=K{V}^{2}t$

Hence, the work done is $K{V}^{2}t$