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A 40 kg slab rests on a frictionless flo...

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 = 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
Limited friction between block and slab `=mu_(s)m_(A)g=0.6 xx 10 xx 10 =60 N`
But applied force on block on A is 100 N.So tha block will slip over a slab.
Now kinetic friction works between block and slab `F_(k)=mu_(k)m_(A)g=0.4 xx 10 xx 10 =40 N`
This kinetic friction helps to move tha slab
` :. "Acceleration of slab" = 40/m_(B)=40/40=1 m/s^(2)`
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