Work, Kinetic Energy and Potential Energy
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CHAPTER 6. WORK, KINETIC ENERGY AND POTENTIAL ENERGY
f = 110
(a) Block moves 5.00 m up plane while acted upon by gravity, friction and an applied force.
(b) Directions of the displacement and the force of gravity.
is pulled 5.00 m. (a) How much work is done by gravity? (b) How much energy
is lost due to friction? (c) How much work is done by the 100 N force? (d) What
is the change in kinetic energy of the crate? (e) What is the speed of the crate
after being pulled 5.00 m?
(a) We can calculate the work done by gravity in two ways. First, we can use the definition:
W = F · d. The magnitude of the gravity force is
= mg = (10.0 kg(9.80
) = 98.0 N
and the displacement has magnitude 5.00 m. We see from geometry (see Fig. 6.4(b)) that the
angle between the force and displacement vectors is 110
. Then the work done by gravity is
= F d cos = (98.0 N)(5.00 m) cos 110
= -168 J .
Another way to work the problem is to plug the right values into Eq. 6.10. From simple
geometry we see that the change in height of the crate was
y = (5.00 m) sin 20
= +1.71 m
Then the work done by gravity was
= -mgy = -(10.0 kg)(9.80
)(1.71 m) = -168 J
(b) To find the work done by friction, we need to know the force of friction. The forces
on the block are shown in Fig. 6.5(a). As we have seen before, the normal force between
the slope and the block is mg cos (with = 20
) so as to cancel the normal component of
the force of gravity. Then the force of kinetic friction on the block points down the slope
(opposite the motion) and has magnitude
N = µmg cos
= (0.400)(10.0 kg)(9.80
) cos 20
= 36.8 N
From simple geometry we see that the change in height of the crate was y = (5.00 m) sin 20 = +1.71 m Then the work done by gravity was W grav = -mgy = -(10.0 kg)(9.80 m s 2 )(1.71 m) = -168 J (b) To find the work done by friction, we need to know the force of friction.