A 2 kg block is pushed up an inclined plane of inclination 37° imparting it a speed of 20ms-1.how much distance will tje block travel before coming to rest?the coefficient of kinetic friction between the block and the incline plane is coefficient of kinetic friction is 0.5.
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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....
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....
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