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If in pervious problem, an additional force `F' = 100N` is applied in vertical direction as shown in figure The friction force acting on the block is

A

`zero`

B

`10 N`

C

`20 N`

D

`5 N`

Text Solution

Verified by Experts

The correct Answer is:
d
Here normal reaction `N = F' = 100 N`
`(F' + mg - N= mg)`
Hence `f_(lim) = mu_(s)N = 0.1 xx 100 = 10N`
Hence block will not move friction will be static
Hence `f = 5N`
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