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

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

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)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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