Question

Determine the time which the smaller block reaches other end of bigger block in the figure:

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

Answer 1

The correct option is C 2.19 s

Given,

Mass of A block $M=8kg$, Mass of B block $m=2kg$

Coefficient of friction, $\mu =0.3$

Friction force $F_r=\mu mg=0.3\times 2\times 10=6N$.

Acceleration of A block $a_A=\frac{F_r}{M}=\frac{6}{8}=0.75m/s^2$

Net force on B block $F_{net}=F-F_r=10-6=4N$

Acceleration of B block $a_B=\frac{F_{net}}{m}=\frac{4}{2}=2m/s^2$

Relative acceleration of block B with respect of A, $a_r=a_B-a_A=2-0.75=1.25m/s^2$

Apply kinematic equation of motion

$s=ut+\frac{1}{2}{a}_r{t}^2$

$3=0+\frac{1}{2}\times 1.25\times t^2$

$t=2.19\sec$

Time require to reach to the other end of the bigger block is $2.19sec$.