A block of mass 1kg is placed on a rough incline as shown. The coefficient of friction between block and incline is 0.4. The acceleration of block is [g=10ms−2,√3=1.7]
A
Zero
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B
1.6ms−2
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C
6.5ms−2
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D
5ms−2
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Solution
The correct option is B1.6ms−2 Force on the block = mgsinθ =1×10×12=5N
Limiting friction = μmgcosθ =0.4×1×10×√32 =2√3 ≈3.4N, as √3=1.7 given F>fL (body slips)
Net force ⇒ma=mgsinθ−μmgcosθ ⇒5−3.4=a a=1.6m/s2