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Question

A 40kg slab rests on a frictionless floor. A 10kg block rests on top of the slab. The coefficient of friction between the block and the slab is 0.40. The 10kg block is acted upon by a horizontal force of 100N. If g=10m/s2, the resulting acceleration of the slab will be
1021215_73f644451f734acaa94fb7282f4d05ac.png

A
1.0m/s2
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B
1.47m/s2
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C
1.52m/s2
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D
6.1m/s2
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Solution

The correct option is A 1.0m/s2
  • Maximum value of friction between slab and block is μN as block has no motion in vertical direction so normal reaction N=mg=10g=10×10=100newton
  • Thus maximum friction is 0.4×100Newton=40Newton
  • This friction is the only horizontal force which act on the slab so acceleration of Slab = friction40Kg= 1 ms2
  • Option A is correct.
  • Note if we treat slab and block both as a system then common acceleration is 100N10Kg+40Kg which is 2ms2 but friction is only able to produce 1ms2 so block will slip over slab.

958254_1021215_ans_6d6f4411e74a4705b1532862a17cacd7.png

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