A crate of mass 9.8 kg is pulled up a rough incline with an initial speed of 1.6

Question

A crate of mass 9.8 kg is pulled up a rough incline with an initial speed of 1.60 m/s. The pulling force is 102 N parallel to the incline, which makes an angle of 20.4° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.94 m.

(a) How much work is done by the gravitational force on the crate?
J

(b) Determine the increase in internal energy of the crate–incline system owing to friction.
J

(c) How much work is done by the 102-N force on the crate?
J

(d) What is the change in kinetic energy of the crate?
J

(e) What is the speed of the crate after being pulled 4.94 m?
m

Explanation / Answer

a)
distance pulled along the incline,d = 4.94 m
rise in height ,h = d* sin 20.4 = 4.94 * sin 20.4 =1.722 m
Work done by gravity = F*h
= -m*g*h
= -9.8*9.8*1.722
= -165.4 J
b)
work done by friction = f*d
= -miu*m*g*cos 20.4 * 4.94
= -0.4*9.8*9.8*cos 20.4 * 4.94
= -177.9 J
So, this must be increase in internal energy due to friction = 177.9 J
c)
work done by pulling force = F*d
= 102*4.94
=503.9 J
use:
kinetic energy at bottom + work done by pulling force = kinetic energy at top + increase in gravitational potential energy + increase in internal energy due to friction
KE (top) - KE (bottom) =work done by pulling force - increase in gravitational potential energy - increase in internal energy due to friction
= 503.9 -165.4 - 177.9
= 160.6 J
Manitude of change in kinetic energy = 160.6 J
d)
KE (top) - KE (bottom) = 160.6 J
0.5*m*(vf^2 - vi^2) = 160.6 J
0.5*9.8*(vf^2 - 1.6^2) = 160.6
vf^2 - 1.6^2 = 32.78
vf = 5.94 m/s
Answer: 5.94 m/s

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