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Question

A 4 kg block is placed on top of a long 12 kg block, which is accelerating along a smooth horizontal table at a=5.2 m/s2 under application of an external constant force. let minimum co-efficient of friction between the two blocks which will prevent the 4 kg block from sliding is μ and co-efficient of friction between the blocks is only half of this minimum value i.e., μ2. Find the amount of heat generated due to sliding between the two blocks during the time in which 12 kg block moves 10 m starting from rest.



A
32 J
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B
42 J
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C
52 J
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D
62 J
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E
32 J
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F
42 J
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G
52 J
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H
62 J
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Solution

The correct options are
C 52 J
G 52 J
First let us assume that both the blocks are moving with common accelaration i.e., a=5.2 m/s2


For 4 kg block:

f=ma

μ0 mg=m (5.2)

μ0=0.52

Given, μ=μ02=12×0.52=0.26,

The acceleration of 4 kg block is due to friction.

a1=μg=2.6 m/s2

As there is relative motion between the blocks, we apply Srel,

Srel=urel+12arel.t2

arel=a1a=2.65.2=2.6 m/s2

Srel=12×(2.6)×t2 ......(1)

Time of motion can be determined from motion of lower block.

For 12 kg block,

S=12×(5.2)×t2

10=12×(5.2)×t2

t2 s

Substituting t in eq(1) we get,

Srel5 m

Work done by friction is given by,

Wf=μ mg Srel=0.26×4×10×(5)

Wf=52 J

Heat generated = 52 J

Hence , option (c) is the correct answer.

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