Question

A block of mass $m=5$ kg is attached with a mass-less spring of force constant $k$. The block is placed over a fixed rough inclined surface for which the coefficient of friction is $0.75$. The block of mass $m$ is initially at rest. The block of mass $M$ is released from rest with spring in un-deformed state. The minimum value of $M$ required to move the block up the plane is (neglect mass of string and pulley and friction in pulley.)

• A. 3 kg
• B. 4 kg
• C. 10 kg
• D. 5 kg

Answer 1

The correct option is A 3 kg

As long as the block of mass m remains stationary, the block of mass M released from rest comes down by $\frac{2Mg}{K}$ (before coming it rest momentarily again).

Thus the maximum extension in spring is $x=\frac{2Mg}{K}$---------(1) for block of mass m to just move up the incline $kx=mgsin\theta +\mu mgcos\theta$-----------(2)

$2Mg=mg\times \frac{3}{5}+\frac{3}{4}mg\times \frac{4}{5}$ or $M=\frac{3}{5}m=\frac{6}{5}mg=3Kg$