Asked by clueless
Starting from rest, a 10.2 kg block slides 2.90 m down a frictionless ramp (inclined at 30.0° from the floor) to the bottom. The block then slides an additional 4.80 m along the floor before coming to a stop. Determine the coefficient of kinetic friction between block and floor.
Answers
Answered by
bobpursley
ap?
the block has mgh of PE when it starts, so at the bottom all that energy is kinetic.
Now, friction absorbs this energy, so
mu*mg*distance=mgh
solve for mu
Notice the distance is independent of mass. Neat experiment to show this.
the block has mgh of PE when it starts, so at the bottom all that energy is kinetic.
Now, friction absorbs this energy, so
mu*mg*distance=mgh
solve for mu
Notice the distance is independent of mass. Neat experiment to show this.
Answered by
Damon
distance down = 2.9 sin 30 = 1.45 meter
Ke at bottom = U at top = m g h = m*g*1.45 Joules
Work done by friction = that KE
mu m g * 4.8
so
m*g*1.45 = mu m g *4.8
mu = 1.45/4.8
note m g cancels
Ke at bottom = U at top = m g h = m*g*1.45 Joules
Work done by friction = that KE
mu m g * 4.8
so
m*g*1.45 = mu m g *4.8
mu = 1.45/4.8
note m g cancels
Answered by
Walter White
I am the one who nocks just put that in
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