physics
posted by angel on .
A 2.0kg wooden block slides down an inclined plane 1.0 m high and 3.0 m long. The block starts from rest at the top of the plane. The coefficient of kinetic friction between the block and plane is 0.35. What is the speed of the block when it reaches the bottom?

Sin¤= 1/3
¤=19.47'
Fnet= Fg//  Ff
Fnet= mg.sin¤ Ff
=2(9.8)sin19.470.35
Fnet= 6.183N
a= Fnet/m
a= 6.183/2
a=3.091m/s^2
(Vf^2)=(Vi^2) +2as
Vf^2= (0)+2(3.091)(3)
Vf^2= 18.546
Vf= 4.31m/s downward.
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how can you solve it using work power and energy?