Question

A force of $30$N acts on a body of mass $2.0$kg starting from rest up to a distance of $3.0$m. Then, the force reduces to $15$N and acts in the same direction up to $2.0$m. Calculate the final kinetic energy of the body.

Answer 1

For the total 3m distance: $F=ma$

$a=\frac{F}{m}=30/2=15m/s^2$

$v^2=u^2+2as$

$=0+2\times 15\times 3$

$v^2=90$

$v=\sqrt{90}m/s$

For the next 2m distance,

$a^{\prime }=\frac{F^{\prime }}{m}=\frac{15}{2}=7.5m/s^2$

$v^{\prime 2}=v^2+2as$

$=90+2\times 7.5\times 2=90+30=120$

Final kinetic energy $=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times 2\times 120=120J$