Question

A $2kg$ block is kept over a rough ground with coefficient of friction as shown in figure. A time varying force F= 2t (F in newton and t in second) is applied on the block. Which of the following statement is correct?

• A. $a=0$ $for$ $t\le 8s,$ $a=t-8$ $for$ $t>8s,$
• B. $a=0$ $for$ $t\le 4s,$ $a=t-8$ $for$ $t>4s,$
• C. $a=0$ $for$ $t\le 8s,$ $a=2t-8$ $for$ $t>8s,$
• D. $a=0$ $for$ $t\le 0.8s,$ $a=t-8$ $for$ $t>0.8s,$

Answer 1

The correct option is A $a=0$ $for$ $t\le 8s,$ $a=t-8$ $for$ $t>8s,$

$\text{Step 1: Free body diagram [Refer Fig.]}$

From F.B.D. $N=mg$

$\text{Step 2: Time Calculation}$

The acceleration will be zero up to a certain time $t$.

The time variable force reaches value of maximum static frictional force just before it is about to move.

$F=f_{\text{Smax}}$ $={\mu }_{s}N={\mu }_{s}mg$

$\Rightarrow 2t={\mu }_{s}mg$

$\Rightarrow 2t=0.8\times 2\times 10$

$t=8s$

$\text{Step 3: Acceleration Calculation}$

The block will start sliding just after $t=8s$ with acceleration $a$

Applying Newton's second law

$\sum F_x=ma$

$\Rightarrow F-f=ma$

$\Rightarrow 2t-0.8\times 2\times 10=2a$

$\Rightarrow a=t-8$

$\therefore$ $a=0$ $for$ $t\le 8s,$ $a=t-8$ $for$ $t>8s,$

Hence Option $A$ is correct.