A 40kg slab (B) rests on a smooth floor. A 10kg b

Question

A 40kg slab (B) rests on a smooth floor. A 10kg block (A) rests on the top of the slab. The static coefficient of friction between slab and block is 0.6 while the kinetic friction coefficient is 0.4.The block A is acted upon by a horizontal force 100N. If g=9.8m/s^2, the resulting acceleration of slab B will be

Arun · Accepted Answer

Normal reaction from 40 kg slab on 10 kg block = 10 * 9.81 = 98.1 N
Static frictional force = 98.1 * 0.6 N is less than 100 N applied
=> 10 kg blck will slide on 40 kg slab and net force on it
= 100 N - kinetic friction
= 100 - 98.1 * 0.4 = 61 N
=> 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s

Frictional force on 40 kg slab by 10 kg block = 98.1 * 0.4 = 39 N
=> 40 kg slab will move with
39/40 m/s
= 0.98 m/s.

Modern Physics> A 40kg slab (B) rests on a smooth floor. ...

Manjeet , 6 Years ago

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Modern Physics> A 40kg slab (B) rests on a smooth floor. ...

Manjeet , 6 Years ago

Arun

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