 # 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.

11 years ago

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...

Fb = mgsin@

total force in downward direction is F+f

F(total) = umgcos@ + mgsin@

manet = mg(ucos@+sin@)

anet = 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....

11 years ago

thnku......bt dis answer is not given in the options... 11 years ago

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@