A block of mass 10kg is placed on a rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of magnitude 100N is applied on the block, then acceleration of the block will be [Takeg=10ms−2]
A
10ms−2
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B
5ms−2
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C
15ms−2
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D
0.5ms−2
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Solution
The correct option is B5ms−2 The said condition can be depicted as shown below
Here it is given that m=10kg,g=10ms−2,μ=0.5andF=100N ∴ Force of friction, f=μN=μmg ⇒f=0.5×10×10=50N
Force that produces acceleration, F′=F−f=100N−50N=50N
Thus, acceleration of the block, a=F′m=5010=5ms−2