Oregon State University
Physics 201
,
Fall 2008
HW8
10 (due Dec. 5 at 5:00 p.m.)
Page 1
Oregon State University
Physics 201
Fall Term, 2008
HW810 Solutions
1.
A pump is to be built to lift 18.0 kg of water per minute through a height of 3.50 m.
What
output rating (in watts) should the pump motor have?
The pump lifts 18/60 kg of water each
second.
That is, it lifts (18/60)(9.80) N, or 2.94 N of water per second.
So:
The force is 2.94 N; the
distance through which that force acts is 3.50 m; and this is done in 1 second.
So the work done per
second is (2.94)(3.50) = 10.3 J.
10.3 J/s =
10.3 W
.
The motorʼs output rating must be 10.3 W.
2.
Chapter 6, problem 24.
Analysis of the work being done:
W
Fk
=
k
F
k
s
cos
θ
=
μ
k
F
N
d·
cos
θ
= –
μ
k
F
N
d
F
N
=
mg
cos
φ
The fundamental WorkEnergy equation:
ME
0
+
W
NC
=
ME
KE
0trans
+
KE
0rot
+
PE
0
+
W
NC
=
KE
trans
+
KE
rot
+
PE
Detailed expansion of the WorkEnergy equation for this situation:
( /
2
)
mv
0
2
+ 0 +
mg(H + d
sin
φ
)
μ
k
mgd
cos
φ
=
( /
2
)
mv
2
+ 0 +
mg
(0)
Simplify (since v
0
= 0):
mg
(
H
+
d
sin
φ
)
μ
k
mgd
cos
φ
=
( /
2
)
mv
2
Divide by m:
g
(
H
+
d
sin
φ
)
μ
k
gd
cos
φ
=
(
1
/
2
)
v
2
Multiply by 2
t
:
2
g
(
H
+
d
sin
φ
) – 2
μ
k
gd
cos
φ
=
v
2
Take the sqare root:
[2
g
(
H
+
d
sin
φ
–
μ
k
d
cos
φ
)]
1/2
=
v
The numbers
:
{2(9.80)[3.50 + (10.4)sin25 – (0.200)(10.4) cos25]}
1/2
=
10.9 m/s
H
μ
k
φ
Initial point (at rest)
Fin
al point (at impact)
d
h = 0
F
k
θ
s
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View Full DocumentOregon State University
Physics 201
,
Fall 2008
HW8
10 (due Dec. 5 at 5:00 p.m.)
Page 2
3.
A certain rollercoaster in California has a circular “looptheloop”
feature with a radius of 8.00 m.
It is designed so that riders are in
“freefall” for the moment when they are at the top of the loop.
What speed must the roller coaster have as it enters the loop and
begins to climb?
(Assume no friction in the track.)
Analyze the situation from an initial point at the bottom of the
loop, where the roller coaster is just entering it (at speed
v
0
), to
the top of the loop, where the roller coaster is moving at (tangential) speed
v
.
For the purposes of measuring
PE
, let
h
= 0 at the bottom of the loop.
The fact that the roller coaster is in freefall at the top indicates that its weight is the only yforce
(i.e. the normal force by the track is momentarily zero), so its weight is equal to the net yforce—
the centripetal force that keeps the roller coaster moving in a circular path:
mg
=
mv
2
/
r
Simplifying this, we have
g
=
v
2
/
r
.
Solving for
v
2
(
v
2
=
rg
), we can substitute it into the above result:
v
0
2
=
rg
+ 4
rg
Collect terms:
v
0
2
=
5
rg
Take the square root:
v
0
=
(5
rg
)
1/2
The numbers:
v
0
=
[5(8.00)(9.80)]
1/2
=
19.8 m/s
8.00 m
Analysis of the work being done:
There is no work being done by any force except gravity (so
W
NC
= 0). Note that there is a non
servative forces acting—the normal force by the track on the roller coaster—but it is always at
right angles to the displacement and hence contributes no work.
But this note about the rotational energy involved here:
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 Fall '08
 Staff
 Physics, Energy, Force, Kinetic Energy, Wnc

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