A 6.00–kg block is set into motion up an inclined plane with an initial speed of
ID: 1776963 • Letter: A
Question
A 6.00–kg block is set into motion up an inclined plane with an initial speed of vi = 8.60 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of = 30.0° to the horizontal.
(a) For this motion, determine the change in the block's kinetic energy.
Recall that the change in a quantity is defined as the final value minus the initial value. J
(b) For this motion, determine the change in potential energy of the block–Earth system.
J
(c) Determine the friction force exerted on the block (assumed to be constant).
N
(d) What is the coefficient of kinetic friction?
Explanation / Answer
a) change in the block's kinetic energy = 1/2 m vf2 - 1/2 m vi2
= 0 - [1/2 * 6.00 * 8.602] = -221.88 J
b) change in potential energy of the block–Earth system = m g h = m g d sin theta
= 6.00 * 9.8 * 3.00 * sin 30 = 88.2 J
c) energy loss = 221.88 - 88.2 = 133.68 J
friction force = E / d = 133.68 / 3.00
= 44.56 N
d) coefficient of kinetic friction = f / m g cos theta
= 44.56 / (6 * 9.8 * cos 30)
coefficient of kinetic friction = 0.875
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