A 5.90-kg block is set into motion up an inclined plane with an initial speed of v₁ = 8.30 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 0 = 30.0° to the horizontal. V; (a) For this motion, determine the change in the block's kinetic energy. (b) For this motion, determine the change in potential energy of the block-Earth system. (c) Determine the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?

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Chapter8: Conservation Of Energy
Section: Chapter Questions
Problem 8.23P: A 5.00-kg block is set into motion up an inclined plane with an initial speed of i = 8.00 m/s (Fig....
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8.7

=
8.30 m/s (see figure below).
A 5.90-kg block is set into motion up an inclined plane with an initial speed of vi
The block comes to rest after traveling d 3.00 m along the plane, which is inclined at an angle of 0 30.0° to
the horizontal.
=
V;
0
(a) For this motion, determine the change in the block's kinetic energy.
(b) For this motion, determine the change in potential energy of the block-Earth system.
(c) Determine the friction force exerted on the block (assumed to be constant).
(d) What is the coefficient of kinetic friction?
Transcribed Image Text:= 8.30 m/s (see figure below). A 5.90-kg block is set into motion up an inclined plane with an initial speed of vi The block comes to rest after traveling d 3.00 m along the plane, which is inclined at an angle of 0 30.0° to the horizontal. = V; 0 (a) For this motion, determine the change in the block's kinetic energy. (b) For this motion, determine the change in potential energy of the block-Earth system. (c) Determine the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?
Part 4 of 6 - Analyze
(b) The change in gravitational potential energy is given by
AU = mgyf - 0
= 5.9
Part 5 of 6 - Analyze
(c) The nonisolated system energy model can be described using the following equation.
AK + AU
-Σω - fkd
other
fk
5.9 kg)(9.80 m/s²)(3
The force of friction is the only other unknown force, so we find it from
AK - AU
d
203
+
X
Your response differs significantly from the correct answer. Rework your solution from the beginning and c
3
3
and, performing the calculations, we find that
116.3
J
m
86.7
= 38.76
3 m sin 30.0°= 86.73
m
N.
86.7 J.
Transcribed Image Text:Part 4 of 6 - Analyze (b) The change in gravitational potential energy is given by AU = mgyf - 0 = 5.9 Part 5 of 6 - Analyze (c) The nonisolated system energy model can be described using the following equation. AK + AU -Σω - fkd other fk 5.9 kg)(9.80 m/s²)(3 The force of friction is the only other unknown force, so we find it from AK - AU d 203 + X Your response differs significantly from the correct answer. Rework your solution from the beginning and c 3 3 and, performing the calculations, we find that 116.3 J m 86.7 = 38.76 3 m sin 30.0°= 86.73 m N. 86.7 J.
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