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A block is released from top of a smooth inclined plane.It reaches the bottom of the plane in 6 sec.The time taken by the body to cover the first half of the inclined plane is:

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A block is released from top of a smooth inclined plane It reaches the bottom of the plane in sqrt2s The time taken (in second) by the body to cover Ist half of inclined plane is .

A block is released from the top of the smooth inclined plane of height h . Find the speed of the block as it reaches the bottom of the plane.

Knowledge Check

  • A block of mass m is released from the top of a mooth inclined plane of height h its speed at the bottom of the plane is proportion to

    A
    `m^(0)`
    B
    `m`
    C
    `m^(2)`
    D
    `m^(-1)`

  • A block released from rest from the top of a smooth inclined plane of angle theta_1 reaches the bottom in time t_1 . The same block released from rest from the top of another smooth inclined plane of angle theta_2 reaches the bottom in time t_2 If the two inclined planes have the same height, the relation between t_1 and t_2 is

    A
    `t_2/t_1=((sin theta_1)/(sin theta_2))^(1//2)`
    B
    `t_2/t_1=1`
    C
    `t_2/t_1=(sin theta_1)/(sin theta_2)`
    D
    `t_2/t_1 =(sin^2 theta_1)/(sin^2 theta_2)`

  • A block released from rest from the top of a smooth inclined plane of angle theta_1 reaches the bottom in time t_1 . The same block released from rest from the top of another smooth inclined plane of angle theta_2 reaches the bottom in time t_2 .If the two inclined planes have the same height the relation between t_1 and t_2 is

    A
    `(t_2)/(t_1) =(( sin theta_1)/( sin theta_2))^(1//2)`
    B
    `(t_2)/(t_1)=1`
    C
    `(t_2)/(t_1 )= ( sin theta_1)/( sin theta_2)`
    D
    `(t_2)/(t_1 ) = ( sin^2 theta_1)/(sin^2 theta_2)`

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