Question

As shown in the figure, there is no friction between the horizontal surface and the lower block $(M=3\, kg )$ but friction coefficient between both the blocks is $0.2$. Both the blocks move together with initial speed $V$ towards the spring, compresses it and due to the force exerted by the spring, moves in the reverse direction of the initial motion. Find the maximum value of $V$ (in $cm / s$ ) so that during the motion, there is no slipping between the blocks. (use $\left.g=10\, m / s ^{2}\right)$.

Answer 1

Maximum chance of slipping occurs when spring is maximum compressed. At this moment, as force exerted by the spring is maximum, acceleration of the system is maximum.
Hence maximum friction force is required at this moment.
By W/E theorem
$\frac{1}{2}(M+m) V^{2}=\frac{1}{2} k x_{m}^{2}$
$\Rightarrow x_{m}=\sqrt{\frac{(M+m) V^{2}}{K}}$
Now for upper block $a_{m}=\frac{k x_{m}}{M+m}$
Force on upper block is provided by the friction force. Therefore
$\mu m g \geq \frac{k x_{m} \cdot m}{M+m}$
For limiting value $V=\mu g \sqrt{\frac{M+m}{k}}$
Using values $V_{\text {maximum }}=20\, cm / s$