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body with zero initial velocity moves down an inclined plane from a height hand then ascends along the same plane with an initial velocity such that it stops at the ame height h. In which case is the time of motion longer?

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Knowledge Check

  • The velocity of a sphere rolling down an inclined plane of height h at an inclination theta with the horizontal, will be :

    A
    `(gh)/(1+k^(2)//r^(2))`
    B
    `(2gh)/(1+k^(2)//r^(2))`
    C
    `sqrt((gh)/(1+k^(2)//r^(2)))`
    D
    `sqrt((2gh)/(1+k^(2)//r^(2)))`
  • A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be

    A
    `sqrt(2 gh)`
    B
    `sqrt(7/10 gh)`
    C
    `sqrt(10/7 gh)`
    D
    `(sqrt(10/7))gh`
  • 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
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    A body with zero initial velocity slips from the top of an inclined plane forming an angle alpha with the horizontal. The coefficient of friction mu between the body and the plane increases with the distance s from the top according to the law mu = bs . After what distance the body will stop.

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