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3. We take +y to be up for both the monkey and the package. The force the monkey pulls
downward on the rope has magnitude F. According to Newtons third law, the rope pulls
upward on the monkey with a force of the same magnitude, so N
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3. Assume in a moment block 1 contacts the spring, having speed v10 , and block 2 has speed
v20 . If v10 > v20 , the spring will be compressed. In contrast, if v10 < v20 , the spring will
be extended. So if the
1. It is easy to know that only gravity does the work on this object. That is we have
K = U
where K and U are for kinetic energy and gravitational potential respectively. More
detailedly, (reference point is at the rotational axis)
1
K = I 2 = U = (M + m)
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3. The free-body diagram for the ball is shown as Fig1. T~1 is the tension exerted by the
upper string on the ball, T~2 is the tension force of the lower string, and m is the mass of
the ball. Note that the tens
1. Because v0 is bigger than R0 , the friction between the ball and the lane accelerates the
angular speed and decelerates the centroid speed of the ball. We can write down two
equations. (Rotational interia of the solid ball is I = 25 mR2 )
Rmgk
5gk
=
I
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4.
1
2L
+1
2 L
F~
m~g
In the figure above, we can write down the equilibrium equation,
mg = 2F cos = m =
2F cos
g
Now we need to find out the critical force, which breaks the net. With the constant volume
of the
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5. The initial and final kinetic energies are zero, and we set up energy conservation in the
form of Eq K = Wgravitation + Wf riction (with K = 0) according to our assumptions.
Certainly, it can only come to a p
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3. We solve the problem using the work-kinetic energy theorem, which states that the change
in kinetic energy is equal to the work done by the applied force, K = W . In our
problem, the work done is W = F d , where F is the tension i
Fundamental of Magnetism Take Home Exam
Due: 2016.04.08
[Problem 1]
Considering a Helmholtz coil, which is arranged by two identical circular loops with
radius R. The current flows in the two coils are both I. The distance between this two
coils is R.
(a)
General Physics (b) (I)
HW1 Solution
1.
(a) v (t)
d
d
x (t ) et t 3 17t 2 29t et 3t 2 34t 29
dt
dt
d
d2
d
v (t ) 2 x(t ) et 3t 2 34t 29 et 6t 34
(b) a(t)
dt
dt
dt
t
2
(c) v (t 3) e 3t 34t 29
t 3
e3 27 102 29 e3 46
2.
a
32 52 ( 4) 2 50 5 2
(a)
(b)
a i
1.
~a
f~
F~
m
M
Assume the maximum amplitude is xmax . In the figure above, we can write down the
motion equation,
Fmax = kxmax = (m + M )amax = amax =
kxmax
m+M
Thus the maximum static friction force is
fmax = mg = mamax = m
m+M
kxmax
= xmax =
g
m+M
k
Sc