A box with an initial speed of #6 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #2/5 # and an incline of #( 5 pi )/12 #. How far along the ramp will the box go?

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Answer 1 · 2017-07-12T06:41:33

Ans: 1.9 m

Given:
u = 6m/s.
#mu# =2/5 = 0.6
#theta =( 5 pi)/12= 75 degrees#

For an inclined plane,

#a = g(sin theta - mu cos theta)#
#a= 9.8 ( sin 75- 0.4 * cos75)#
#a= 8.45 m/(s^2)#

Again, #v= u -( asin theta)t#
#u/(asintheta) = t#
#t= 0.634 s#

Thus,
#s= ut - 1/2 gsintheta * t^2#

Therefore, # s= 1.9 m#

Thank you!!