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


A 40 kg slb rest on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static corfficient 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 100N. If `g=9.8(m)/(s^2)`, the resulting acceleration of the slab will be.

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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