Question

The mass of a body is 2.5 kg. It is in motion with velocity (v), after time (t) $v=\frac{t}{3}+\frac{t^2}{2}+1$. Calculate force acting at t = 3 sec and 10 sec.

Answer 1

We know that by differentiating $\frac{dv}{dt}=a$ (acceleration)

Given evaluation $v=\frac{t}{3}+\frac{I^2}{2}+1$

Force acting at $t=3sec$

$a=\frac{dV}{dt}=\frac{1}{3}+\frac{2t}{3}=\frac{1}{3}+3=\frac{10}{3}m/s^{2}F=ma=\frac{2.5}{10}\times \frac{10}{3}=\frac{25}{3}N$

(mass$=2.5kg$ given)

Force acting at $t=10sec$

$a=\frac{1}{3}+10=\frac{31}{3}m/s^{2}F=ma=\frac{2.5}{10}\times \frac{31}{3}=\frac{775}{30}N$