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A block of mass m sliding down an inclin...

A block of mass `m` sliding down an incline plane. The incline plane has fixed base length `'l'` and coefficient of friction on the incline plane is `'mu'`. The plane is fixed and block slides from top to bottom. Let `theta_(0)` be the inclination angle for minimum sliding time and `v_(0)` be the block's speed when it reaches the bottom in that case. Pick the correct option (s):

A

`v_(0)=sqrt(2gl(tantheta_(0)-mu))`

B

`v_(0)=sqrt(2gl tantheta_(0))`

C

`tan(2theta_(0))=(-1)/(mu)`

D

`tan(theta_(0))=1/(mu)`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`t^(2)=(2l)/(g[cos alpha sin alpha -mu cos^(2) alpha])`
For min. time, `d/(dalpha)(cosalpha sin alpha-mucos^(2)alpha)=0`
`implies tan 2alpha =(-1)/(mu)`
`/_\K+/_\U=W_("friction")`
`impliesv_(0)=sqrt(2gl(tantheta_(0)-mu))`
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