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Describe what happens to a box attached to a pulley + spring

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Homework Statement


The figure below shows a block attached to a pulley and a string.

1) which quantity has essentially the same nonzero value at all three points?

2) which quantity changes direction at point Z?

3)which quantity is zero at point Y?


Homework Equations


Restoring force = -kx, where k is constant and x is the distance from equilibrium.

Force of gravity = mg, where m is mass and g is acceleration due to gravity.

KE=½mv^2, where m is mass and v is velocity.

Velocity= distance over time

Net force equal mass times acceleration.


The Attempt at a Solution


1) Force of gravity acting on the block is the same, since mass didn’t change and g is always constant. D then.

I have a question here.Because Y is midway, does that mean it’s the equilibrium position?

2) I doubt my answer here. KE is a scalar, so no change in direction. Force of gravity is always downward. That leaves A, B, and E.

The force on the string is the restoring force that is always pulling the mass to its equilibrium position. So, if it is at point Z the restoring force pulls it upward to its equilibrium position. If it’s at point X, restoring force pulls it downward to its restoring position.

For velocity, the object is changing its direction when it oscillates between X and Z. Not sure what happens to direction of the acceleration. All I know is that, at equilibrium position, velocity is maximum while acceleration is zero, but for direction I guess it changes direction as the velocity? But I chose E.

3) I learned that at maximum distance, X and Z, the restoring force and the acceleration are maximum while velocity is zero. At equilibrium position, velocity is maximum while acceleration and restoring force are zero. If Y is the equilibrium position, there’s two possible answers, A and E. I thought of the acceleration of the system as a nonzero constant. If Y is not the equilibrium position, so it is B. Confusing..
 

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Answers and Replies

  • #2
BvU
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Hi,

1) I agree
2) It appears strange to me that the wire should pull downward at e.g. point X. (Taut) Wires can only pull in one direction. Consider trying again ...
3) You say velocity is zero at X and at Z. Can you imagine the block as not moving at Y ? Consider again ...

The real way out for you may be to make a graph of position vs time and draw in the five quantities.
 
  • #3
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Hi,

1) I agree
2) It appears strange to me that the wire should pull downward at e.g. point X. (Taut) Wires can only pull in one direction. Consider trying again ...
3) You say velocity is zero at X and at Z. Can you imagine the block as not moving at Y ? Consider again ...

The real way out for you may be to make a graph of position vs time and draw in the five quantities.
I treated the system as a simple harmonic motion. Wouldn’t that be a sine wave?

Ah, it doesn’t pull downward. But it like loosens a bit.. as it moves upward, the spring compresses. As it moves downward the spring stretches
 
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  • #4
CWatters
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1) Correct
2) Incorrect. I've yet to see a string that can push an object.
3) I believe Y is meant to be the equilibrium position. You gave A and E as possible answers. E is incorrect. A is correct.

Edit: Cross posted with BvU.
 
  • #5
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1) Correct
2) Incorrect. I've yet to see a string that can push an object.
3) I believe Y is meant to be the equilibrium position. You gave A and E as possible answers. E is incorrect. A is correct.

Edit: Cross posted with BvU.
Is the force on the string different from the restoring force? Yes, the string pulls upward, but wouldn’t the spring stretch and the box move to Z.
Can you tell me why is E incorrect?
 
  • #6
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2) considering your reply, I have to choose between A and B. Although I still can’t get why the force on the string would not move up at Z and down at X. But B seems reasonable, too. Changing the direction of motion changes the velocity. But what about A?
 
  • #7
CWatters
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Best thing you can do is what BvU said and make a plot of position, velocity and acceleration. Yes they look like sin waves but they aren't in phase with each other.

The restoring force in this problem is the sum of forces provided by the spring/string and gravity. In position X gravity > string tension. In position z string tension > gravity.
 
  • #8
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Best thing you can do is what BvU said and make a plot of position, velocity and acceleration. Yes they look like sin waves but they aren't in phase with each other.

The restoring force in this problem is the sum of forces provided by the spring/string and gravity. In position X gravity > string tension. In position z string tension > gravity.
Aha, I have to take into account the force of gravity.
I graphed position vs time graph already, and have found that acceleration is zero at X and Z while the velocity is momentarily constant. Next, velocity is increasing at Y while acceleration is constant. Velocity changes the direction at X and Y. Right?
 
  • #9
CWatters
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Sorry most of that is wrong.

The acceleration isn't zero at X and Z.

The velocity isn't increasing at Y. The acceleration at Y isn't constant.

The velocity changes direction at X and Z (not Y) but perhaps that's a typo?
 
  • #10
CWatters
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Post your plots?
 
  • #11
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Hmm...okay the velocity is zero at maximum distance,X and Z, while acceleration here has maximum value.

At Y, there’s no acceleration but velocity has its maximum value.

The velocity changes direction at X and Z (not Y) but perhaps that's a typo?
Yes yes, that was a typo.

Not a good graph, but maybe the right one?
 

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  • #12
CWatters
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Think velocity graph starts ok but after half a cycle gets a bit out of alignment with the position graph.
 
  • #13
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Got that. But the quantities are correct?
 
  • #14
CWatters
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Hmm...okay the velocity is zero at maximum distance,X and Z, while acceleration here has maximum value.

At Y, there’s no acceleration but velocity has its maximum value.
Yes that's all ok.
 
  • #15
BvU
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Today is Friday - a bit late, perhaps ?

I see position starting at X for t=0 and then going down. That can never correspond to a positive velocity in the lower graph ...

Tip: draw acceleration too and add some vertical dotted lines spanning all three plots, for mental alignment :rolleyes:
 
  • #16
CWatters
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You're right. Should be increasing but in the negative direction.
 

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