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A block of 4 kg slides down an inclined ...

A block of `4 kg` slides down an inclined plane which makes an angle `theta = 60^(@)` with the horizontal. The co-effcient of friction between the block and the plane is `mu = 3//4`. The force exerted by the block on the plane is `(g = 10 m//s^(2))`

A

`25N`

B

`125//4N`

C

`15N`

D

`75//4 N`

Text Solution

Verified by Experts

The correct Answer is:
A
A block ……………
Resultant force `= N sqrt(1+mu^(2))`
`= mg cos theta sqrt(1+mu^(2))`
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