Question

The minimum value of coefficient of friction (u) such that block of mass '5 kg' remains at rest is [NCERT Pg. 102] 5 kg 3 kg (1) 0.3 (3) 0.6 (2) 0.5 (4) 0.4​

Answer 1

Answer:

The minimum value of coefficient of friction (u) such that block of mass ‘5 kg’ remains at rest.

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.)

As long as the block of mass m remains stationary, the block of mass M released from rest comes down by

K

2Mg

(before coming it rest momentarily again).

Thus the maximum extension in spring is x=

K

2Mg

---------(1) for block of mass m to just move up the incline kx=mgsinθ+μmgcosθ-----------(2)

2Mg=mg×

5

3

+

4

3

mg×

5

4

or M=

5

3

m=

5

6

mg=3KgA cube of 10N rests on plane of coefficient of friction 0.6. The slope of the plane is 3 in 5. The minimum force required to start the cube moving up the plane is

A body slides down a rough inclined plane of inclination θ with constant velocity v. If it is projected up the plane with velocity 2v, then it moves up the plane .

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