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Work Energy Theorem, Kinetic Energy, and Tension

by IAmSparticus
Tags: kinetic energy, tension, work energy theorem
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IAmSparticus
#1
Oct16-09, 10:28 PM
P: 36
1. A rescue helicopter lifts a 90 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance of 10 m. What is the tension in the cable and how much work is done by the tension in the cable? Use the work-energy theorem to find the final speed of the person as well.


2. Work Energy theorem: Wnet = (1/2 mass velocity final^2) - (1/2 mass velocity initial^2)
Tension: T = mass gravity



3. Tension: mass gravity
T = 90 kg * 9.8 m/s^2
T = 882 N
Which is wrong according to Webassign.

WET = Delta K
WET = (1/2 mass velocity final^2) - (1/2 mass velocity initial^2)
Not sure how to go about solving this part. I guess the masses cancel, but what next?
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cepheid
#2
Oct16-09, 10:35 PM
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Quote Quote by IAmSparticus View Post
3. Tension: mass gravity
T = 90 kg * 9.8 m/s^2
T = 882 N
It's no surprise that that is wrong. Remember that the person is accelerating upwards. This means that there is a net upward force on him. If the tension is only just balancing his weight, then obviously the net force is zero, and he won't be accelerating. If that's not clear, then you need to review Newton's 2nd Law.
IAmSparticus
#3
Oct16-09, 10:44 PM
P: 36
Ok, so I tried using Newtons Second Law, Force = mass * acceleration, and now I got 63 N, which is apparently still wrong.

cepheid
#4
Oct16-09, 10:48 PM
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Work Energy Theorem, Kinetic Energy, and Tension

Quote Quote by IAmSparticus View Post
Ok, so I tried using Newtons Second Law, Force = mass * acceleration, and now I got 63 N, which is apparently still wrong.
Yeah, that's the NET force which is pulling him upwards. Draw a free body diagram for the person, and you'll see that two forces are acting on him, namely his weight downwards, and the force due to the tension in the cable, upwards. You add the two forces acting on him together (taking direction into account) to get the net (total) force on the person. The upward force (tension) must **exceed** his weight by 63 N in order for there to be a NET upward force of that amount accelerating him upward. Do you understand?
IAmSparticus
#5
Oct16-09, 10:58 PM
P: 36
I think I understand, so would the net force be -9.1 m/s^2 * 90 kg which would equal -819 N?
cepheid
#6
Oct16-09, 11:09 PM
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No. Did you draw a free body diagram for the person like I suggested?

T is pulling him upwards.

Weight is pulling him downwards. Let's call the magnitude of the weight W. Then the weight is -W.

To find the net force, add the two forces that are acting on him:

T + (-W) = T - W = Fnet = +63 N

For T - W to be positive, T must be greater than W. In other words, the force pulling him upwards is greater than the force pulling him downwards, so that the force pulling him upwards "wins." This is what we mean when we say that there is a NET upward force on him. "Net" = the direction and magnitude of the end result, after all forces have been considered.

T - W = 63 N.

You know what W is.
IAmSparticus
#7
Oct16-09, 11:13 PM
P: 36
W is equal to the weight, which is the gravitational force times the mass, so -9.8 m/s^2 * 90 kg... which is -882 N. Correct?
cepheid
#8
Oct16-09, 11:22 PM
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Quote Quote by IAmSparticus View Post
W is equal to the weight, which is the gravitational force times the mass, so -9.8 m/s^2 * 90 kg... which is -882 N. Correct?
Yeah, that's correct. Now, as I've stated before, the strength of the force pulling him upwards must be greater than that (in magnitude) by 63 N.
IAmSparticus
#9
Oct16-09, 11:30 PM
P: 36
So I add 63 N to -882 N to get -819 N? Or is it positive 819 N since it is in the opposite direction of the weight? Or would it be 945 N since that is what 63 N + 882N (the opposite of the weight) equals?
cepheid
#10
Oct16-09, 11:36 PM
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There is only one of those possibilities that makes sense (the third one that you wrote). I am going to tell you the reason for the fourth time: the force pulling him upward has to be larger (in magnitude) than the force pulling him downward in order for him to be moving upward.

One way to think of it:

T - W = 63

T - 882 = 63

T = 882 + 63

Just think of it in terms of magnitudes (ignoring direction). The *magnitude* of the weight is 882 N, and the magnitude of the tension is 63 N larger than that.
IAmSparticus
#11
Oct16-09, 11:40 PM
P: 36
So the work done by the tension would be equal to the force times the displacement, or 945 N * 10 m, which would equal 9450 J?
cepheid
#12
Oct16-09, 11:52 PM
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Quote Quote by IAmSparticus View Post
So the work done by the tension would be equal to the force times the displacement, or 945 N * 10 m, which would equal 9450 J?
Sounds about right to me!
IAmSparticus
#13
Oct17-09, 12:03 AM
P: 36
So then to find the final speed of the person I'm supposed to use the work energy theorem which will help me find the change in kinetic energy which is [ (1/2 mass velocity final^2) - (1/2 mass velocity initial^2) ]?


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