Question

A particle of mass $50g$ moves on a straight line. The variation of speed with time is shown in figure. Find the force acting on the particle at $t=2,4$ and $6$ seconds.

Answer 1

$m=50g=5\times {10}^{-2}kg$

As shown in the figure.

Slope of $OA=\tan \theta \frac{AD}{OD}=\frac{15}{3}=5m/s^2$

So, at $t=2sec$ acceleration is $5m/s^2$

Force $=ma=5\times {10}^{-2}\times 5=0.25N$ along the motion.

At $t=4\sec$

slope of $AB=0$, acceleration $=0[\tan 0^o=0]$

$\therefore$ Force $=0$

At $t=6sec$, acceleration = slope of $BC$

In $\Delta BEC=\tan \theta =\frac{BE}{EC}=\frac{15}{5}=5$.

Slope of $BC=\tan ({180}^o-\theta )=-\tan \theta =-5m/s^2$ (deceleration)

Force $=ma=5\times {10}^{-2}5=0.25N$. Opposite to the motion.