A 5.40–kg block is set into motion up an inclined plane with an initial speed of
ID: 1498473 • Letter: A
Question
A 5.40–kg block is set into motion up an inclined plane with an initial speed of vi = 7.60 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.
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).
You do not have a coefficient of kinetic friction, so you cannot use Fk = kn. What affect does the friction force have on the total mechanical energy? N
(d) What is the coefficient of kinetic friction?
Explanation / Answer
The change in the block's kinetic energy is.
del KE = 1/2 m ( vf^2- vi^2)
= 1/2 *5.4 ( 0-7.6^2)
=-155.952 J
(b)
del PE = mg d sin theta = 5.4 ( 9.8)3 sin 30 =79.38 J
(c)
the energy loss due to the friction is
E = KE - PE = 155.952 J- 79.38 J=76.52 J
E= f d
f = E/d = 76.52/3=25.52 N
(d)
fk = uk mg cos theta
uk = fk/ mg cos thets
= 25.5N/ ( 5.4) 9.8* cos 30
=0.556
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