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# Work, Kinetic Energy and Potential Energy

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6.2. WORKED EXAMPLES
137
f
fric
F
app
d
F
app
f
fric
mg
N
(a)
(b)
Figure 6.3:
(a) Applied force and friction force both do work on the box. (b) Diagram showing all the
forces acting on the box.
The vertical forces acting on the box are gravity (mg, downward) and the floor's normal
force (N, upward). It follows that N = mg and so the magnitude of the friction force is
f
fric
= µN = µmg = (0.30)(40 kg)(9.80
m
s
2
) = 1.2 × 10
2
N
The friction force is directed opposite the direction of motion ( = 180
) and so the work
that it does is
W
fric
= F d cos
= f
fric
d cos 180
= (1.2 × 10
2
N)(5.0 m)(-1) = -5.9 × 10
2
J
or we might say that 5.9 × 10
2
J is lost to friction.
(c) Since the normal force and gravity do no work on the box as it moves, the net work done
is
W
net
= W
app
+ W
fric
= 6.5 × 10
2
J - 5.9 × 10
2
J = 62 J .
By the work­Kinetic Energy Theorem, this is equal to the change in kinetic energy of the
box:
K = K
f
- K
i
= W
net
= 62 J .
(d) Here, the initial kinetic energy K
i
was zero because the box was initially at rest. So we
have K
f
= 62 J. From the definition of kinetic energy, K =
1
2
mv
2
, we get the final speed of
the box:
v
2
f
=
2K
f
m
=
2(62 J)
(40 kg)
= 3.1
m
2
s
2
so that
v
f
= 1.8
m
s
8. A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of
1.50
m
s
. The pulling force is 100 N parallel to the incline, which makes an angle of
20.0
with the horizontal. The coefficient of kinetic friction is 0.400, and the crate

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Summary :

## It follows that N = mg and so the magnitude of the friction force is f fric = µN = µmg = (0.30)(40 kg)(9.80 m s 2 ) = 1.2 × 10 2 N The friction force is directed opposite the direction of motion ( = 180 ) and so the work that it does is W fric = F d cos = f fric d cos 180 = (1.2 × 10 2 N)(5.0 m)(-1) = -5.9 × 10 2 J or we might say that 5.9 × 10 2 J is lost to friction.

Tags : box,force,fric,friction,work,energy,kinetic,app,net,cos,180,incline,graity
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