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HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 1 of 6
a).
For Spring 2, Spring
force: F=k*
∆
x,
∆
x=0.60.1 =0.5 (m), Spring Force = object weight = 3 (N)
k = F/
∆
x =3/0.5= 6 (N/m)
b.)
For Spring 3,
F= k*
∆
x,
∆
x=1.10.1 =1.0 (m),
Object weight= Spring Force= k*
∆
x= 6* 1=6 (N)
In orbit, he has centripetal acceleration, so the total net force along the radius direction pointing to the earth
center is NOT zero. Hence if he stands on a scale, the normal force from the scale is not equal to mg. Also when
he is in orbit, g is NOT equal to 9.8 m/s
2
a).
A mass on spring: Oscillation period:
,
So:
So:
=3000*1
2
/
(4*3.14
2
)=76
kg
b). Potential energy stored in the spring
U =1/2* k*
∆
x
2
When it is Furthest from the equilibrium length:
∆
x=Amplitude=4m
U =1/2* k*
Α
2
=0.5*3000*4
2
=24000 J
c). When it passes equilibrium: All Potential energy is converted into Kinetic energy:
½*m*V
2
=U
max
= 1/2* k*
Α
2
=24000 J
V
2
=k*A
2
/m
d). oscillation period is in dependant to amplitude. It is only determined by the stiffness of the spring (spring
constant) and the mass of the object. So T is still 1 s.
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View Full DocumentHW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 2 of 6
Problem 3.
The height of a tower is measured by attaching a simple pendulum to its
ceiling, whose length is barely enough to stay off the floor. The pendulum is let go
from a small angle, and takes
12
s to return to the same location it started from.
a.)
How tall is the tower?
m
If the pendulum mass is let go
0.2
m above the floor,
b.)
how fast is the mass traveling as it grazes the floor?
m/s
Solution.
(a)
The period of the pendulum is related to the length of the
pendulum,
H=L
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 Spring '09
 JamesM.Lockhart
 Physics, Force

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