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A 40kg slab rest on a frictionless floor...

A 40kg slab rest on a frictionless floor as shown in . A 10kg blocks rests on the top of the slab. The static coefficient of friction between the block and the slab is `0.60`, while kinetic friction is `0.40`. The 10kg block is acted upon by a horizontal force of 100N. What is the resulting acceleration of the slab ?

A

`1(m)/(s^2)`

B

`1.5(m)/(s^2)`

C

`2(m)/(s^2)`

D

`6(m)/(s^2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Limiting friction between block and slab`=mu_Sm_Sg`
`=0.6xx10xx10=60N`
But applied force on block A in 100 N, so the block will slip over a slab.
Now kinetic friction works between block and slab `f_k=mu_km_Ag`
`=0.4xx10xx10=40N`
This kinetic friction helphs to move the slab
Acceleration of slab `=(40)/(m_B)=(40)/(40)=1(m)/(s^2)`
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