Question

A body of mass $2kg$ rests on a horizontal plane having coefficient of friction $\mu =0.5$. At $t=0$ a horizontal force $F$ is applied that varies with time $F=2t$. The time $t_0$ at which motion starts and distance moved in $t=2t_0$ second will be ______and ______respectively.

• A. $3sec,25/6m$
• B. $5sec,25/6m$
• C. $3sec,125/6m$
• D. $5sec,125/6m$

Answer 1

The correct option is D $5sec,125/6m$

Maximum static friction = $\mu N=0.5\times 2g=10N$
Thus, Force $F=10N$
$\Rightarrow 2t=10$
$\Rightarrow t=5=t_0$ seconds, at which it starts moving (when maximum static friction is reached)
Given $F=2t$, means $2a=2t$ or $a=t$ or $\frac{dv}{dt}=t$, which on integrating gives $v=t^2/2=\frac{ds}{dt}$
Thus, integrate again to get $s=\frac{t^3}{6}+constant$
At $t=0$, $s=0$. Hence, $constant=0$
Now, for interval, $2t_0=10$ seconds,it starts moving at $t=5$ seconds. It moves a distance of $\frac{5^3}{6}=\frac{125}{6}m$