A block of mass 'm' is projected up an inclined pane of inclination θ with an initial velocity 'u'. If the coefficient of kinetic friction between the block and the pane is μ, the distance upto which the block will rise up the plane , before coming to rest, is given by (a) u²μ/2gsinθ (b) u²μ/2gcosθ (c) u²/4gsinθ (d) u²/4gcosθ .Please explain.

friction force acts on the block in backward direction (down the plane),

f = uN                 (N is normal reaction)

N = mgcos@

f = umgcos@ 

one of the component of weight acts in backward direction & other perpendicular to plane...

F_b = mgsin@

total force in downward direction is F+f

 F_(total) = umgcos@ + mgsin@

 ma_net = mg(ucos@+sin@)

   a_net = g(ucos@+sin@)      .............1

now  , we can use , v² = u² - 2as                ( coz accleration is constant)

            finally v becomes 0 , so

    s = u²/2a

 plugging value of a from eq 1

  s = u²/2g(sin@+ucos@)

this is value of distance covered before stopping....

thnku......bt dis answer is not given in the options...

may be they have taken some assumptions or ur q is not complete ....

can u tel me the ans of this q??

answer given is  u²/4gsinθ.

your q was in complete , this is the q & sol ....earlier this q was asked by kaushik in this forum..

Q=> A block slides down an inclined plane of slope angle  θ with constant velocity. It is then projected up the same plane with an initial velocity 'u'  plane of inclination  . How far up the incline will it move before coming to rest?

solution =>

lock moves down the plane with constant velocity it means net force is zero...

force due to friction (f) = mgsin@                              (@ is the angle of plane)

net retardation force when body is projected upward is (f+mgsin@)

 f = mgsin@ so

net retardation force = 2mgsin@

     ma = mgsin@

     a(retardation) = 2gsin@        .............1

 V² = U² - 2aS

finally velocity becomes 0 so

  U² = 2aS

 S = U²/2a

    =u²/4gsin@