A 30kg block rests on a rough horizontal surface. A horizontal force of 200N is applied on the body. The block acquires a speed of 4m/sec, starting from rest, in 2sec. What is the value of coefficient of friction?
A
10√3
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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 C0.47
So equation for the motion is, F−fk=ma where fk=μR=μmg=μ×30×9.8=294μ Since initial velocity of the body is zero & its velocity increases from zero to 4m/sec in 2 seconds, so acceleration of the body is given by, v=u+at ⇒4=0+a×2 or a=2m/sec2 So, F−fk=30×2 ⇒200−μ×294=60 ⇒μ=140294=0.476