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Question

A 40 kg slab rests on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force 100 N. If g equals 9.8 m divided by s to the power of 2 end exponent, the resulting acceleration of the slab will be

  1. 0.98 m divided by s to the power of 2 end exponent    
  2. 1.47 m divided by s to the power of 2 end exponent    
  3. 1.52 m divided by s to the power of 2 end exponent    
  4. 6.1 m divided by s to the power of 2 end exponent    

The correct answer is: 0.98 m divided by s to the power of 2 end exponent


    Limiting friction between block and slabequals mu subscript s end subscript m subscript A end subscript g
    equals 0.6 cross times 10 cross times 9.8 equals 58.8 N
    But applied force on block A is 100 N. So the block will slip over a slab.
    Now kinetic friction works between block and slab F subscript k end subscript equals mu subscript k end subscript m subscript A end subscript g equals 0.4 cross times 10 cross times 9.8 equals 39.2   N
    This kinetic friction helps to move the slab
    therefore Acceleration of slabequals fraction numerator 39.2 over denominator m subscript B end subscript end fraction equals fraction numerator 39.2 over denominator 40 end fraction equals 0.98   m divided by s to the power of 2 end exponent

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