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A block is projected upwards with a velo...

A block is projected upwards with a velocity of 5 m/s from the bottom of a rough inclined plane of inclination 30° with the horizontal. It stops after 0.5 s. What is the coefficient of friction ? (g`=10m//s^2`)

A

`sqrt3`

B

`1/(sqrt3)`

C

`sqrt2`

D

`sqrt5`

Text Solution

Verified by Experts

The correct Answer is:
B
When the block moves up, the frictional force acts downward and the resultant downward
Force=`mgsintheta+mumgcostheta`
This is a retarding force.
`:.` Acceleration down the plane,
`a=F/m=g(sintheta+mucostheta)`
For the upward motion of the block,
v=u+at
where v=0, u=5 m/s, t=0.5s and a negative
`:." "0=5-10(sin30^(@)+mucos30^(@))xx0.5`
`:." "0=5-5xx1/2-(5mu*sqrt3)/(2)`
`:." "5/2=5/2sqrt3mu" ":." "sqrt3mu=1" or "mu=1/(sqrt3)`
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