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Question

A 40kg slab rests on frictionless floor. A 10 kg block rests on the top of the slab. The stafic coefficient of friction between block and the slab is 0.60 while the Kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g=9.8 m/s2 find the resulting accel. Of the slab (0.98 m/s2)

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Solution

Normal reaction from 40kg slab on 10kg block = 10×9.81=98.1N

Static frictional force = 98.1×0.6N is less than 100 N applied

= 10 kg block will slide on 40 kg slab and net force on it

= 100 N - kinetic friction

= 10098.1×0.4=61N

=> 10 kg block will slide on 40 kg slab with 6110 = 6.1m/s

Frictional force on 40 kg slab by 10 kg block = 98.1×0.4=39N

=> 40 kg slab will move with

3940m/s

= 0.98m/s.

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