Laws Of Motion Question 396

Question: 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 $ \mu =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.)

Options:

A) $ \frac{3}{5}m $

B) $ \frac{4}{5}m $

C) $ \frac{6}{5}m $

D) $ \frac{3}{2}m $

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Answer:

Correct Answer: A

Solution:

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 the spring is $ x=\frac{2Mg}{K} $ …(1)

For block of mass m to just move up the incline $ kx=mg\sin \theta +\mu mg\cos \theta $ …..(2)

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



NCERT Chapter Video Solution

Dual Pane