Question

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

Answer 1

2.19 s

Explanation:

1) We have,

Let us make free body diagram of 2 kg block.

Normal Reaction on 2kg block,

N-2g=0 ( No Vertical motion )

Maximum value of friction on 2kg block,  

$f_{max}=\mu N=\mu*mg=0.3*2*10=6 N$

2) As 10 N > 6N , 2kg block will move relative to 8 kg.

Again, FBD of 2kg Block.

$10-6=2*a\\ \\=> a= 2m/s^2$

3) But, we need to find relative acceleration of 2kg block.

So, FBD of 8 kg block.

That is,

$f_r=8a\\ \\=>6 = 8a\\ \\=> a=\frac{3}{4} m/s^2$

Now, Relative acceleration of 2kg block is given by

$a_{rel}=2-\frac{3}{4}=\frac{5}{3}$

Hence,

$S_{rel}=u_{rel}*t+\frac{1}{2}a_{rel}t^2\\ \\=>3=0+\frac{1}{2}*\frac{5}{4}t^2\\ \\=>t=\sqrt{\frac{24}{5}}=2.19 s$

Hence, Required time is 2.19 s