Energy and Momentum [Spring (word problem)]

In summary, the homework statement asks for the work done when an object is compressed by a given amount, but does not provide information about the maximum compression.
  • #1
AlexisFair
13
0

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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  • #2
Looks fine to me. The question is not very clear but I agree you should assume that the spring is compressed 11 cm and then let go with the object attached to the end.

AM
 
  • #3
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.
 
  • #4
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.
 

1. What is the equation for calculating potential energy in a spring?

The equation for calculating potential energy in a spring is U = 1/2kx2, where k is the spring constant and x is the displacement from the equilibrium position.

2. How does the mass of an object affect its kinetic energy in a spring system?

The mass of an object does not affect its kinetic energy in a spring system. The kinetic energy is solely determined by the velocity of the object.

3. Can potential energy be negative in a spring system?

Yes, potential energy can be negative in a spring system. This occurs when the spring is compressed and the displacement x is negative. It is important to note that the total energy (potential + kinetic) remains constant.

4. How does the spring constant affect the energy stored in a spring?

The spring constant affects the energy stored in a spring by determining how much force is needed to stretch or compress the spring. A higher spring constant means a stiffer spring, which requires more force to stretch or compress, resulting in more energy stored.

5. Can the energy of a spring system be converted into other forms of energy?

Yes, the energy in a spring system can be converted into other forms of energy, such as thermal energy or sound energy. This occurs when the spring is released and the potential energy is converted into kinetic energy, which can then be transferred to other objects or transformed into other forms of energy.

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