Energy and Momentum [Spring (word problem)]

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AlexisFair
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Homework Statement

Homework4.jpg
[/B]

Please check the attachments as to view the questions.

Homework Equations



Ek= 1/2 mv^2

EPSpring= 1/2 kx^2

The Attempt at a Solution



A) EPSpring= 1/2 kx^2

1/2 (100)(0.11) ^2
=0.605 JB) Is ok I understand it.

C)

EKO=EKFEPSpring(Initial) + EK(Initial) = EPSpring(Final) + EK(Final)
½ kx^2 + ½ mv^2 = ½ kx^2 + ½ mv^20.605 J + 0 J = ½ (100) (0.055) ^2 + ½ (0.065) v^2V= 3.74 m/s

Is this correct? Also, for EK(Initial) it is 0 J because it is being "compressed" 11 cm and thus it has no speed, correct?
 
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Part c is so unclear you should not be penalised for giving the wrong answer.

I think they mean it's compressed 11cm, then released, and they want to know the velocity after the object has traveled 5.5cm.

5.5cm is half of 11cm so it does not matter if the 5.5cm is measured from the initial rest position or the compressed position.
 
This is a truly poorly constructed problem.

1. The problem doesn't specify whether the force is applied for some time (how long?) and then removed, or whether it remains in effect. Either will result in SHM, so that's not diagnostic. Why do they say that the force is exerted on the spring, rather than on the object? It's a strange assertion even if it proves to be inconsequential.

2. They ask about the work done compressing the spring by a given amount, but they don't tell you that this is the maximum compression so it leaves you guessing as to whether or not it informs you about the SHM in any way.

3. The speed of the object at a given location in its travels will depend on whether or not the applied force was constant or removed (the equilibrium location depends upon this) and you still don't know if you're working with the maximum compression or not from part (a).

The best you can do is to make and state assumptions that don't contradict what is given, then answer the questions based upon those assumptions. You can't be faulted if your assumptions are clear and justifiable.