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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) u 2 μ/2gsinθ (b) u 2 μ/2gcosθ (c) u 2 /4gsinθ (d) u 2 /4gcosθ .Please explain.

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.

Grade:12

5 Answers

vikas askiitian expert
509 Points
10 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....

Aiswarya Ram Gupta
35 Points
10 years ago

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

vikas askiitian expert
509 Points
10 years ago

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

can u tel me the ans of this q??

Aiswarya Ram Gupta
35 Points
10 years ago

answer given is  u2/4gsinθ.

vikas askiitian expert
509 Points
10 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@

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