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A block slides down a rough inclined pla...

A block slides down a rough inclined plane of slope angle `theta` with a constant velocity. It is then projected up the same plane with an initial velocity v. The distance travelled by the block up the plane before coming to rest is

A

`(v^(2))/(4g sin theta)`

B

`(v^(2))/(2g sin theta)`

C

`(v^(2))/(g sin theta)`

D

`(4gv^(2))/(sin theta)`

Text Solution

Verified by Experts

The correct Answer is:
A
`a_(d)=g(sin theta-mu_(k)costheta)`
`a_(u)=g(sin theta+mu_(k)cos theta), v^(2)-u^(2)=2as`
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Knowledge Check

  • A block slides down a rough inclined plane of slope angle theta with a constnat velocity. It is then projected up the same plane with an intial velocity v the distance travelled by the block up the plane coming to rest is .

    A
    `(v^(2))/(4sintheta)`
    B
    `(v^(2))/(2sintheta)`
    C
    `(v^(2))/(gsintheta)`
    D
    `(4gv^(2))/(sintheta)`

  • A block can slides down an inclined plane of slope and theta , with constant velocity. If it is projected up the same plane with an initial speed v_(0) , then

    A
    The distance moved up the incline before coming to rest `(v_(0)^(2))/(2 g sin theta)`
    B
    The distance moved up the incline before coming to rest `(v_(0)^(2))/(4 g sin theta)`
    C
    Block will remain at rest after reaching maximum height
    D
    Block will moving down with uniform velocity after reaching maximum height

  • A block of mass m slides down an inclined plane which makes an angle theta with the horizontal. The coefficient of friction between the block and the plane is mu . The force exerted by the block on the plane is

    A
    `mg cos theta`
    B
    `sqrt(mu^(2) + 1) mg cos theta`
    C
    `(mumg cos theta)/(sqrt(mu^(2) + 1))`
    D
    `mumg theta`