10.14.
Model:
Assume an ideal spring that obeys Hooke’s law.
Visualize:
Solve:
(a)
The spring force on the 2.0 kg mass is
sp
.
F
k y
=  ∆
Notice that
y
∆
is negative, so
sp
F
is positive.
This force is equal to
mg
, because the 2.0 kg mass is at rest.
We have
.
k y
mg
 ∆ =
Solving for
k
:
2
(
/
)
(2.0 kg)(9.8 m/s )/( 0.15 m ( 0.10 m))
392 N/m
k
mg
y
= 
∆
= 

 
=
(b)
Again using
:
k y
mg
 ∆ =
2
e
e
/
(3.0 kg)(9.8 m/s )/(392 N/m)
0.075 m
0.075 m
0.10 m
0.075 m
0.175 m
17.5 cm
y
mg k
y
y
y
y
∆ = 
= 
′
′

= 
⇒
=

= 

= 
= 
The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The
position
of the end of the
spring is negative because it is below the origin, but length must be a positive number.
10.34.
Model:
This is case of freefall, so the sum of the kinetic and gravitational potential energy does not
change as the cannon ball falls.
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View Full DocumentVisualize:
The figure shows a beforeandafter pictorial representation. To express the gravitational potential energy, we put
the origin of our coordinate system on the ground below the fortress.
Solve:
Using
y
f
=
0 and
the equation
i
gi
f
gf
K
U
K
U
+
=
+
we get
2
2
2
2
i
i
f
f
i
i
f
2
2
2
f
i
i
1
1
2
2
2
2
(80 m/s)
2(9.8 m/s )(10 m)
81.2 m/s
mv
mgy
mv
mgy
v
gy
v
v
v
gy
+
=
+
⇒
+
=
=
+
=
+
=
Assess:
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 Spring '08
 Nandi
 Physics, Energy, Force, Kinetic Energy, Mass, Potential Energy

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