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Question

A 30 kg block rests on a rough horizontal surface. A horizontal force of 200 N is applied on the body. The block acquires a speed of 4 m/sec, starting from rest, in 2 sec. What is the value of coefficient of friction?

A
103
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B
310
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C
0.47
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D
0.185
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Solution

The correct option is C 0.47

So equation for the motion is,
Ffk=ma
where fk=μR=μmg=μ×30×9.8=294μ
Since initial velocity of the body is zero & its velocity increases from zero to 4 m/sec in 2 seconds, so acceleration of the body is given by,
v=u+at
4=0+a×2 or a=2 m/sec2
So, Ffk=30×2
200μ×294=60
μ=140294=0.476

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