Question

The figure shows a block placed on a bracket. The bracket is placed unconstrained on a smooth floor and is pulled by a constant force $\vec{F}=6\hat{i}$ horizontally. The block is projected with velocity $v_0$ relative to the bracket as shown in the figure. If the time (in second) after which it stops relative to bracket is $t$, then find the value of $0.6t$ Horizontal surface of bracket is smooth while vertical surface is rough. (Given: $m=1kg,M=5kg,v_0=5m/s,\mu =0.5$ )

Answer 1

Answer: 5

$F=(m+M)a_x$
$\Rightarrow a_x=\frac{6}{5+1}=1m/s^2$
$N_x(\text{ on }m)=m{a}_x=1\times 1=1N$
Acceleration of block,
$a_1=\frac{\mu N_x}{m}(-\hat{k})=\frac{0.5\times 1(-\hat{k})}{1}=0.5(-\hat{k})m/s^2$
Acceleration of bracket,
$a_2=\frac{\mu N_x}{M}(\hat{k})=\frac{0.5\times 1}{5}(\hat{k})=0.1\hat{k}m/s^2$
Relative acceleration,
$a_{\text{rel }}=a_1-a_2=0.6(-\hat{k})m/s^2$
$0=v_0-a_{\text{rel }}t$
$t=\frac{5}{0.6}s$