A 5.40–kg block is set into motion up an inclined plane with an initial speed of
ID: 287973 • Letter: A
Question
A 5.40–kg block is set into motion up an inclined plane with an initial speed of vi = 7.40 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.
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) delta KE = - 1/2 m v2
= -0.5 * 5.40 * 7.402
change in the block's kE = -147.8 J
b) delta PE = m g d sin theta
= 5.40 * 9.8 * 3.00 * sin 30º
change in potential energy = 79.38 J
c) Work done = |delta KE| - |delta PE|
= 147.8 - 79.38 = 68.42 J
friction force = E / d = 68.42 / 3.00
friction force exerted on the block = 22.8 N
d) mu = f / N = f / m g cos theta
= 22.8 / (5.4 * 9.8 * cos 30)
coefficient of kinetic friction = 0.498
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