Question

Determine the time in which the smaller block reaches other end of bigger block in Fig.

• A. 4 s
• B. 8 s
• C. 2.19 s
• D. 2.13 s

Answer 1

The correct option is C 2.19 s

Refer image .1

$f_{smax}={\mu }_{s}mg$

$=\left(0.3\right)\left(2\right)\left(10\right)$

$f_{smax}=6N$

$\therefore$ The block would move,

for $2kg$,

$10-f=2a$

$\Rightarrow 10-6=2a$

$\Rightarrow a=2m/s^2$

for $8kg$

Refer image 2.

$f=m{a}^{\prime }$

$6=8a^{\prime }$

$a^{\prime }=0.75m/s^2$

Now, by relative motions eq.

$S_{net}=U_{net}t+\frac{1}{2}{0}_{net}{t}^2$

$\Rightarrow 3=\frac{1}{2}(2-0.75)t^2$

$\Rightarrow t=\sqrt{\frac{6}{7.25}}$

$t=2.19sec$