 1 
Physics 112  HW #7 Solutions
Fall 2005
6.4
[Sliding Crate]
(a) Crate is sliding at constant speed.
Apply Newton’s 2
nd
Law:
x direction
Σ
F
x
= F
a
– f
k
r
= ma = 0
⇒
F
a
= f
k
r
(1)
y direction
Σ
F
y
= N  mg = 0
(2)
Kinetic friction:
f
k
r
=
µ
k
N
Substitute from (1) and (2):
F
a
=
µ
k
mg = (0.25)(30.0 kg)(9.81 m/s
2
)
=
73.6 N
(b) Work done on crate by applied force
F
a
:
W =
F
a
. s
= (73.6 N)(4.5 m)
=
331 J
(c) Work done on crate by friction
f
k
:
W =
f
k
. s
= (73.6 N)(4.5 m) =
331 J
(d) Work done on crate by normal force and by gravity are both
0 J
, since these forces are
both perpendicular to the displacement.
(e) Total work done on crate is
0 J
.
We know this because crate does not accelerate, i.e. its KE does not change.
6.13
[Alpine Rescue]
Work done on box by friction to travel uphill along slope
by a distance
l
:
W
f
=
f
k
. s
=  f
k
l
(1)
This work is negative because the friction force is in the
opposite direction to the displacement.
Increase in vertical height of box is h :
l
sin
α
= h,
or
l
= h/sin
α
(2)
Applying Newton’s law in direction perpendicular to
slope gives
N = mg cos
α
(3)
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 Fall '07
 LECLAIR,A
 mechanics, Force, Friction, mg cos

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