p07_038.pdf

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38. (a) The force
F
of the incline is a combination of normal and friction force which is serving to “cancel”
the tendency of the box to fall downward (due to its 19.6 N weight). Thus,
F
=
mg
upward. In
this part of the problem, the angle
φ
between the belt and
F
is 80
◦
. From Eq. 7-47, we have
P
=
F v
cos
φ
= (19.6)(0.50) cos 80
◦
which leads to
P
= 1.7 W.
(b) Now the angle between the belt and
F
is 90
◦
, so that
P
= 0.
(c) In this part, the angle between the belt and
F
is 100
◦
, so that
P
= (19.6)(0.50) cos 100
◦
=
−1.7
W.

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