A block of mass 10kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100N is applied on it along the surface , then acceleration of block will be [Take g=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 Here, m=10kg,g=10ms−2,μ=0.5,F=100N
Force of friction, f=μN=μmg =0.5×10kg×10ms−2=50N
Force that produces acceleration F′=F−f=100N−50N=50N a=F′m=50N10kg=5ms−2