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Question

The coefficient of static friction between the block of 2 kg and the table shown in figure is μs = 0.2. what should be the maximum value of m so that the blocks do not move? Take g = 10 m/s2. The string and the pulley are light and smooth.


A

2 kg

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B

1 kg

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C

0.4 kg

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D

0.5 kg

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Solution

The correct option is C

0.4 kg


Consider the equilibrium of the block of mass m. The forces on this block are

(a) mg downward by the earth and (b) T upward by the string.

Hence, T-mg=0 or, T=mg. ...............(i)

Now consider the equillibrium of the 2kg block.This forces on this block are

(a) T towards right by the string.

(b) f towards left (friction) by the table,

(c) 20 N downward (weight) by the earth and

(d) N upward (normal force) by the table,

For vertical equilibrium of this block,

N = 20 N. . . . . . . (II)

As m is the largest mass which can be used without moving the system, the friction is limiting.

Thus, f = μs N. . . . . . . (III)

For horizontal equilibrium of the 2 kg block,

f = T . . . . . . (IV)

Using equations (I), (III) and (IV)

μs N = mg

Or, 0.2 × 20 N = mg

Or, m=0.2×2010kg=0.4kg.


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