A block of mass m is attached with massless spring of force constant k. The block is placed over a fixed rough inclined surface for which the coefficient of friction is $μ = 3 / 4$ . The block of mass m is initially at rest. The block of mass M is released from rest with spring in unstretched 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.)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{3}{5} m$

$\frac{3}{2} m$

$\frac{4}{5} m$

$\frac{6}{5} m$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} As long as the block of mass m remains stationary, the \\ block of mass M released from rest comes down by \\ \frac{2 Mg}{k} (before coming it rest momentarily again). Thus \\ the maximum extension in the spring is \\ x = \frac{2 Mg}{k} \\ For block of mass m to just move up the incline \\ kx = mg \sin θ + μ mg \cos θ \\ \Rightarrow 2 Mg = mg \times \frac{3}{5} + \frac{3}{4} mg \times \frac{4}{5} \\ \Rightarrow M = \frac{3}{5} m \end{array}$