HW Solutions #
6
 8.01 MIT  Prof. Kowalski
Potential Energy and Mechanical Energy Conservation.
NOTE:
We made a mistake on the solution sheet for HW #5.
On part
d
of problem
5
on HW #5 we omitted the work due to
friction. If you submitted the answer
1
2
mv
2
0
x
2
−
x
1
x
1
+
1
2
mv
2
1
it should
have been graded
correct
, and you can recover any lost credit by
resubmitting it to your grader with a note to regrade this problem.
1)
7.42
Please refer to the figure 7.30 p.277.
Let’s denote the compression length as ∆
x
, the velocity after leaving
the spring as
v
0
and the maximum height it goes on the incline as
h
.
Using energy conservation law:
1
2
mv
2
1
+
1
2
kx
2
1
+
mgy
1
+
W
other
=
1
2
mv
2
2
+
1
2
kx
2
2
+
mgy
2
(1)
There is no friction present in this problem and
W
other
= 0. I will
measure the gravitation potential energy with respect to the height
of the dashed horizontal bar indicated in the figure 7.30.
a)
Comparing the energy initially (when spring compressed):
v
1
= 0
x
1
= ∆
x
y
1
= 0
(2)
with the point it leaves the spring:
v
2
=
v
0
x
2
= 0
y
2
= 0
(3)
0 +
1
2
k
∆
x
2
+ 0 =
1
2
mv
2
0
+ 0 + 0
(4)
1
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∴
v
0
=
k
m
∆
x
(5)
b)
For this part we should compare the energy of particle at its
highest point on the incline:
v
3
= 0
x
3
= 0
y
3
=
h
(6)
with the point that it has left the spring:
v
2
=
v
0
x
2
= 0
y
2
= 0
(7)
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 Spring '08
 WHITE
 Physics, Energy, Friction, Potential Energy, Work

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