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 $0.6\,t$ Horizontal surface of bracket is smooth while vertical surface is rough. (Given: $m=1 \,kg , M=5\, kg , v_{0}=5 \,m / s , \mu=0.5$ )

Answer 1

$F=(m+M) a_{x}$
$ \Rightarrow a_{x}=\frac{6}{5+1}=1 m / s ^{2}$
$N_{x}(\text { on } m)=m a_{x}=1 \times 1=1 N$
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 $
$t =\frac{5}{0.6} s$