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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...
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....
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@
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