Solutions to sample exam 2
TAM 212, Fall, 2002
Solution for sample problem #1
1
: The change in kinetic energy of the plunger = work done by external forces:
Work
E
K
=
∆
.
.
(
)
(
)
2
2
2
/
1
2
/
1
.
.
initial
plunger
final
plunger
v
m
v
m
E
K
−
=
∆
0
=
initial
v
, so
(
)
(
)
(
)
2
2
/
32
/
5
2
/
1
.
.
v
s
ft
lb
E
K
=
∆
,
where
v
is its speed just before it hits the button.
The external forces include that due to gravity and that due to the spring.
( )(
)
lb
in
lb
in
h
mg
Work
down
gravity
−
=
−
=
∆
=
75
.
8
75
.
1
5
(
)
(
)
2
2
2
/
1
final
initial
sping
k
Work
δ
−
δ
=
To find the
δ
’s we need to solve for it in the equilibrium position:
(
)
in
lb
in
lb
F
F
ma
F
gravity
spring
y
y
5
.
0
5
/
10
0
=
δ
⇒
=
δ
⇒
=
+
⇒
=
∑
so
(
)(
) (
)
(
)
(
)
lb
in
in
in
in
lb
Work
sping
−
=
−
==
1875
.
2
75
.
0
1
/
10
2
/
1
2
2
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
2
2
/
667
.
11
12
/
1
/
140
1875
.
2
75
.
8
/
/
078125
.
s
ft
in
ft
x
s
ft
in
v
lb
in
v
s
ft
lb
=
=
⇒
−
+
=
⇒
s
ft
v
/
42
.
3
=
⇒
(down)
2
Let m be the mass of the plunger and M be the mass of the button / base combination.
During the collision, external forces are negligible, so conservation of momentum gives:
button
plunger
plunger
Mv
mv
Mx
mv
,
2
,
2
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 Spring '08
 Keane
 Energy, Force, Momentum, Potential Energy, Work, plunger

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