7.9.
I
DENTIFY
:
tot
B
A
W
K
K
. The forces on the rock are gravity, the normal force and friction.
S
ET UP
:
Let
0
y
at point
B
and let
y
be upward.
0.50 m
A
y
R
. The work done by friction is negative;
0.22 J
f
W
.
0
A
K
. The freebody diagram for the rock at point
B
is given in Figure 7.9. The acceleration of
the rock at this point is
2
rad
/
a
v
R
, upward.
E
XECUTE
:
(a)
(i) The normal force is perpendicular to the displacement and does zero work.
(ii)
2
grav
grav,
grav,
(0.20 kg)(9.80 m/s
)(0.50 m)
0.98 J
A
B
A
W
U
U
mgy
.
(b)
tot
grav
0
(
0.22 J)
0.98 J
0.76 J
n
f
W
W
W
W
.
tot
B
A
W
K
K
gives
2
1
tot
2
B
mv
W
.
tot
2
2(0.76 J)
2.8 m/s
0.20 kg
B
W
v
m
.
(c)
Gravity is constant and equal to
mg
.
n
is not constant; it is zero at
A
and not zero at
B
. Therefore,
k
k
f
n
is
also not constant.
(d)
y
y
F
ma
applied to Figure 7.9 gives
rad
n
mg
ma
.
2
2
2
[2.8 m/s]
(0.20 kg)
9.80 m/s
5.1 N
0.50 m
v
n
m g
R
.
E
VALUATE
:
In the absence of friction, the speed of the rock at point
B
would be
2
3.1 m/s
gR
. As the rock
slides through point
B
, the normal force is greater than the weight
2.0 N
mg
of the rock.
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 Fall '07
 Evrard
 Force, Friction, Gravity, Work, Normal Force, kg, m/s, Wtot, Wgrav

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