Question

A 2 kg block is pushed up an inclined plane of inclination 37 ° imparting it a speed of 20 m s-1.how much distance will be the block travel before coming to rest? The coefficient of kinetic friction between the block and the inclined plane is coefficient of kinetic friction is 0.5. Take g = 10 m s-2 and use sin 27°=0.6, cos 37°=0.8

Answer 1

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....i hope tis may help und mark me brilantist