Question

A body of mass 2 kg initially at rest is moved by a horizontal force of 0.5 N in a smooth table. Obtain the work done by the force in 8 s and show that this equals the change in kinetic energy of the body.

Answer 1

Dear Student ,
Here in this case from the given data we can write that ,
$W=Fs=0·5\times 8=4J$
$$\begin{gathered} Nowlook,E=\frac{1}{2}m{v}^2 \\ Now,fromequationofmotion,weget, \\ v=u+at \\ \Rightarrow v-u=at \\ \Rightarrow v=atwhenu=0. \\ Now,E=\frac{1}{2}m{v}^2=\frac{1}{2}m{a}^2{t}^2=\frac{1}{2}{\left(\frac{F}{m}\right)}^2{t}^2=\frac{1}{2}\times {\left(\frac{0·5}{2}\right)}^2\times 64=4J \end{gathered}$$
So the work done is equal to the change in kinetic energy .
Regards