Physics: stumped on this work problem

[FutureScaresMe]

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taken from BR Physics Ch 3 III.11

So it's your standard slide problem. The question asks: "suppose that as a child slides down the slide we take air resistance into account. The work done by the child on the air is equivalent to the change in:
A. KE
B. PE
C. KE + dPE
D. KE - dPE

So, Ei + W on sys = Ef
W on sys = -W by sys
Ei = PE
Ef = KE

Therefore, W by sys = PE - KE
But the book's answer is C. What am i doing wrong?

 

[FutureScaresMe]

So, thinking about it more

I got ahead of myself and assumed KEi = 0 and PEf = 0, but the Q was asking for thr change in KE and PE

(KEi + PEi) + W on sys = (KEf + PEf)
W on sys = dKE + dPE and since the magnitude of work on surr will be the aame and I know it will be positive, since slider is losing E to the air, this is the answer.

This the right way of thinking about it?

 

[MBHockey]

This problem gave me some trouble too. I ultimately just used conservation of energy to get the answer, but my "gut" really doesn't like using equations. I'm going to keep thinking about this one conceptually, but here's an easy way to get the answer (W is work on the air).

Let's apply conservation of energy at the top of the slide and then at a point just after he starts sliding (point 1):


PEi + KEi + Wi = PE1 + KE1 + W1

solve for W1

(PEi-PE1) + (KEi-KE1) = W1

W1 = (change in PE) + (change in KE)

But conceptually, I don't really understand how you can lose potential energy to air resistance or friction since it does not deal with motion. I'm trying to think of a better way to think about it.

 

[Westernhealer]

Conservation of energy is the best way to solve this problem as well as most problems comparing energy before and after motion

 

[Will Hunting]

This problem gave me some trouble too. I ultimately just used conservation of energy to get the answer, but my "gut" really doesn't like using equations. I'm going to keep thinking about this one conceptually, but here's an easy way to get the answer (W is work on the air).

Let's apply conservation of energy at the top of the slide and then at a point just after he starts sliding (point 1):


PEi + KEi + Wi = PE1 + KE1 + W1

solve for W1

(PEi-PE1) + (KEi-KE1) = W1

W1 = (change in PE) + (change in KE)

But conceptually, I don't really understand how you can lose potential energy to air resistance or friction since it does not deal with motion. I'm trying to think of a better way to think about it.

It does deal with motion. Let's assume there is no friction. Then the Work = the loss in potential energy. At top of slide there is potential, but as it gains speed is sacrifices this for kinetic energy. In reality it doesn't do work since gravity is a conservative force. You're simply converting energy. However, when the mcat asks for work then this isn't the case.

Now, let's consider air resistance. Air resistance removes some of the kinetic energy that the child would otherwise have. Remember, total energy is conserved. At the beginning we have potential. In the middle, we have potential and kinetic BUT some of the kinetic was lost due to friction (air resistance). In order for energy to be equal. The GAIN in kinetic combined with the potential change, must equal the work friction done on the slide.

KEi +PEi +Wf=KEf+PEf Air resistance is doing negative work on the child. So, the system is "losing" energy. The equation is for Mechanical Energy; when friction is present, you have to account for that. Lets assume that it started from rest. The equation becomes Pei + Wf= Kef +Pef. The kid has less VELOCITY as a result of air resistance since less of the energy is converted to kinetic energy. Don't think of it as losing POTENTIAL energy. Think of it as losing KINETIC energy. The change in potential is negative and th kinetic is positive; however, the overall sum is negative since wf is negative.

 

[Will Hunting]

This problem gave me some trouble too. I ultimately just used conservation of energy to get the answer, but my "gut" really doesn't like using equations. I'm going to keep thinking about this one conceptually, but here's an easy way to get the answer (W is work on the air).

Let's apply conservation of energy at the top of the slide and then at a point just after he starts sliding (point 1):


PEi + KEi + Wi = PE1 + KE1 + W1

solve for W1

(PEi-PE1) + (KEi-KE1) = W1

W1 = (change in PE) + (change in KE)

But conceptually, I don't really understand how you can lose potential energy to air resistance or friction since it does not deal with motion. I'm trying to think of a better way to think about it.

Essentially you were correct. The correct way is ChangePe +ChangeKe. Ke is positive but PEchange is negative. So you're actually doing Ke + -Pe= Ke-Pe. However, in this question, it is conceptually different than saying Ke-Pe change. You lose PE anytime you fall. It's not due to air resistance. That's just in the nature of things. It's KE that really gets affected in these problems since you can no longer equate change in PE to change in Ke. The question would have been simpler if they asked the work done by air resistance. This is negative work as it acts against the direction of motion.

Finally, I think the question is misleading. There are two forces on the kid. One is gravity and the other is air resistance. The sum of these forces acting over a distance is the overall work. I see what they were getting at. However, resistance acts on the kid, the kid doesn't act on air resistance.

Ex. If a ball is dropped from 100m. With no friction, it's KE at end is 1000 (assuming 1 kg). Now, if we have a resistance of 4 newtons by air. the KE is 600. You see, 400 of negative was done by resistance. However, the question can't say the kid acts on air cuz it doesnt. HTH, i have to go now.

 

[MBHockey]

Essentially you were correct. The correct way is ChangePe +ChangeKe. Ke is positive but PEchange is negative. So you're actually doing Ke + -Pe= Ke-Pe.

Yeah, that's what I just realized. How silly!

I just got some time to sit down and think about this and remained convinced the way I was thinking about it was right. It's just a trivial formality of adding and subtracting.

To the OP, here's what I was just thinking about which made me realize I wasn't crazy:

Let's pretend we have a normal slide, at we're taking into account friction (air resistance, surface to surface friction, whatever, all lumped into one term: Wf)

At the top, let's say some object has 100 Joules of PE and nothing else. Then, let's pick some arbitrary point (point 1) that's a bit down the slide.

Let's pretend that at point 1, the PE is 60 Joules and the KE is 35 joules. What does this tell us? Well, if there were no friction, the change in PE would all go to increasing the KE from zero. So, if there were no friction, KE should be equal to 40 joules. However, we see that we've lost 5 joules somewhere, as in the real scenario the object only has 35 joules of KE (60+35 only equals 95; energy was not conserved). Where did it go? Friction of course. Looking at it this way, it seems the answer should be just change in KE minus change in PE, but the distinction comes from remembering that PE was a negative change, and KE was a positive change.

So at point 1 on the slide,

100 = mgh + 1/2mv^2 + Wf
0 = (60-100) + (35-0) + Wf or

-Wf = delta PE + delta KE

-Wf = -40 +35
Wf = 5 joules

Note: i'm calling Wf positive and just remembering that it is doing negative work. You could easily just change the sign.

hope that's useful.