Speed or turning point using energy

[JessicaHelena]

Homework Statement

Please look at the attached screenshot.
This problem is really confusing for me and I can't seem to make much sense out of it.

Homework Equations

Ei = Ef

The Attempt at a Solution

As you can see, I did get (a). (The other checkmarks, I guessed — there were only two possible choices for each, anyway.) KEi = KEf + 15, and by plugging in appropriate numbers, I was able to get v = 1.9464 m/s.

But as for (b), (c) and (d), I've no idea how to approach them.

I did try (c), and this time using KEi = KEf + 35, I realized there wouldn't be a valid v, and thus proceeded to determine the turning point. Hence my new equation was KEi = mgh, from which I got h=2.1556 m, but that's vertical. We want horizontal distance, however. So using the ratio (1:35/(mg)) = (x:2.1556), x= 0.4707. However, that is wrong.

As for forces, I don't know how to get started.

Please help — I don't know what to do anymore!

 

[jbriggs444]

But as for (b), (c) and (d), I've no idea how to approach them.

Let us work, for the moment, on (b): "What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m?"

As the ball moves leftward from x=4.0 m to x=2.0 m it gains some amount of potential energy. You know a distance covered. You know an energy difference. You want to know a force. Is there a quantity or equation that uses those three things?

 

[DrClaude]

I did try (c), and this time using KEi = KEf + 35, I realized there wouldn't be a valid v, and thus proceeded to determine the turning point. Hence my new equation was KEi = mgh, from which I got h=2.1556 m, but that's vertical. We want horizontal distance, however. So using the ratio (1:35/(mg)) = (x:2.1556), x= 0.4707. However, that is wrong.

Nowhere does it say that the potential energy is due to gravity. You should approach it from the point of view of conservation of energy.

 

[JessicaHelena]

@jbriggs444 — ah, okay. Now that you mention which distances I could use ... It's (35-20)/(4-2) = 7.5 N. Doing the same way, I can get that (d) is 35N. Thank you for that!

 

[JessicaHelena]

@DrClaude — I was trying to use mgh to my advantage here. the change in potential energy in (c) is 55-20 = 35J, and letting that equal mgh, I can get h by doing 35/(mg). Is this wrong?

 

[DrClaude]

@DrClaude — I was trying to use mgh to my advantage here. the change in potential energy in (c) is 55-20 = 35J, and letting that equal mgh, I can get h by doing 35/(mg). Is this wrong?

Nothing in the problem says that the potential energy is due to gravity, much less due to Earth's gravity.

 

[JessicaHelena]

@DrClaude
okay. So initially there's only KE. At the turning point, however, it's no longer moving so all of it should be potential energy. Am I right? Should I be using F*d = W = E somewhere?

 

[jbriggs444]

@DrClaude
okay. So initially there's only KE. At the turning point, however, it's no longer moving so all of it should be potential energy. Am I right? Should I be using F*d = W = E somewhere?

You should be able to do (c) without concerning yourself with force. Concentrate on energy. Yes, you are right that at the turnaround event, its kinetic energy is zero so all of its energy is potential energy.

However, "initially there's only KE" is incorrect. What is the initial PE?

 

[JessicaHelena]

I'm just not sure how I can get a distance just using energy...

 

[jbriggs444]

I'm just not sure how I can get a distance just using energy...

You are given the potential energy at x=6. You are given the potential energy at x=5. You know how much potential energy you have at the turnaround.

This is a simple problem of linear interpolation. Can you find an equation for potential energy in terms of x that is valid between x=5 and x=6? Can you solve for the x that gives the potential energy you know?

 

[CWatters]

The particle starts at a known point with some known KE and PE. As it moves to the right the PE changes according to the graph. COE means it's KE will change as it moves.

Try plotting another line on the graph to illustrate it's KE. Does it cross the KE=0 axis? If so where?

 

[JessicaHelena]

Okay, so Ei = (1/2)(0.78)(6.5)^2 + 20 and Ef is equivalent to that (and is a potential energy). We know that a 6cm PE is 55J, and since we see a linear graph, at x cm (the value we're trying to find at c), it would be Ei = Ef = 36.4775 J.

x/(36.4775) = 6/55, so x = 3.97936 m.

In retrospect, the answer doesn't seem right to me... What did I do wrong?

 

[jbriggs444]

Okay, so Ei = (1/2)(0.78)(6.5)^2 + 20 and Ef is equivalent to that (and is a potential energy). We know that a 6cm PE is 55J, and since we see a linear graph, at x cm (the value we're trying to find at c), it would be Ei = Ef = 36.4775 J.

x/(36.4775) = 6/55, so x = 3.97936 m.

In retrospect, the answer doesn't seem right to me... What did I do wrong?

x=3.97936 could be correct if the graph of potential energy versus position were linear from x=0 through x=6. But it it not. It is linear from x=5 to x=6.

Edit: Oops. That is not your only error.

 

[DrClaude]

x/(36.4775) = 6/55, so x = 3.97936 m.

In retrospect, the answer doesn't seem right to me... What did I do wrong?

You have to figure out the slope, which is not 6/55.

 

[DrClaude]

Edit: Oops. That is not your only error.

Indeed. The particle is traveling to the right, so it should reach a value of x greater than where it started .

 

[jbriggs444]

Indeed. The particle is traveling to the right, so it should reach a value of x greater than where it started .

It is worse than that. I ran the numbers. The assumption that it stops before x=6 is questionable. @JessicaHelena, can you show your work indicating why the particle cannot get past x=6?

 

[JessicaHelena]

The slope is actually 55/1 so I would have x/(36.4775) = 1/55 and x would be the distance from 5. That gives me x= 0.6632 and so the answer would be 5.6632?

 

[jbriggs444]

The slope is actually 55/1 so I would have x/(36.4775) = 1/55 and x would be the distance from 5. That gives me x= 0.6632 and so the answer would be 5.6632?

The slope is not 55/1 either. What is $U_A$? What is $U_C$?

 

[JessicaHelena]

Oh right, 55-20 = 35, so 35/1.

 

[JessicaHelena]

Then would it be 6.04221?

 

[jbriggs444]

Then would it be 6.04221?

But what happens at x=6?

 

[JessicaHelena]

It reaches the maximum potential energy? Is there a problem I don't see yet?

 

[jbriggs444]

It reaches the maximum potential energy?

Right. And what does that mean for the particle's continued motion past that point?

 

[JessicaHelena]

I'm not too sure... the particle will only have KE then.

 

[jbriggs444]

I'm not too sure... the particle will only have KE then.

Will the KE be changing?

Reread the question: "If the particle can reach x=7.0 m, what is its speed there"

 

[JessicaHelena]

No, it'll stay constant. So you think I should get the speed?

 

[jbriggs444]

No, it'll stay constant. So you think I should get the speed?

In your initial post, you indicated that no such speed was possible. It seems that we have demonstrated differently now. So yes, it is time to go back and see if (c) can be answered that way.

 

[JessicaHelena]

But it doesn't seem to make sense because the answer (the one that says m for meters) is already marked correct...

 

[jbriggs444]

But it doesn't seem to make sense because the answer (the one that says m for meters) is already marked correct...

We go for correct answers here. Not answers that we think the answer key thinks are correct.

 

[DrClaude]

The slope is actually 55/1 so I would have x/(36.4775) = 1/55 and x would be the distance from 5. That gives me x= 0.6632 and so the answer would be 5.6632?

Apart from the slope being incorrect, as @jbriggs444 pointed out, you have to be careful about the energy scale. You can work either with the total energy or with the initial state defining the zero of energy. Here, you are mixing both.