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In the figure shown...

In the figure shown

A

1.65 `m//s^2`

B

0.98 `m//s^2`

C

1.2 `m//s^2`

D

0.25 `m//s^2`

Text Solution

Verified by Experts

The correct Answer is:
B

Let the net acceleration of the slab be a limiting friction
`F_s=mumg=0.6xx10xx9.8=58.8N`
`100Ngt58.8N`
I.e. slab will accelerate with different acceleration.

f=40a
`0.4xx10xx9.8=40aimplies a = 0.98 m//s^2`
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