Question

A body of weight W = 120N is pushed by a horizontal for P = t2 + 6. A frictional force f= t3- 5 resists
its motion. If the body starts from rest, find the velocity of the body after 3 seconds.

Answer 1

$F=ma=m\frac{dv}{dt}\to v=\frac{1}{m}\int F(t)dt$

$v=\frac{1}{m}\int (t^2+6-t^3+5)dt=\frac{1}{m}(t^3/3+6t-t^4/4+5t)+C=$

$=\frac{1}{m}(t^3/3+11t-t^4/4)+C$

if $t=0$ $v=0 \to$ $C=0$

We have

$v=\frac{1}{m}(t^3/3+11t-t^4/4)=\frac{1}{120/9.81}(3^3/3+11\cdot 3-3^4/4)=1.78 m/s$