What is the potential energy of a compressed spring on an inclined plane?

In summary, a 2.4 kg block is placed against a spring on a frictionless 17o incline with a spring constant of 59.2 N/cm. The spring is compressed 35.3 cm and released, and the elastic potential energy of the compressed spring is calculated using the formula W = (1/2)kx^2, where k is the spring constant and x is the compression distance. The answer may differ from the book's answer due to incorrect calculations or not taking into account gravitational potential energy. Converting units correctly is also important in solving this problem.
  • #1
scavok
26
0
https://chip.physics.purdue.edu/protected/Halliday6Mimg/h8p19.gif
(ignore values in image)

A 2.4 kg block is placed against a spring on a frictionless 17o incline. The spring, whose spring constant is 59.2 N/cm , is compressed 35.3 cm and then released.

What is the elastic potential energy of the compressed spring in J?


According to my calculations and my book, the potential energy of a spring is .5kx2. That didn't work, so I converted cm to m in k and x. That didn't work either. The only difference between this problem and a similar one in the book is that this spring is at an angle. Could someone give me a hint as to what the trick with this problem is?
 
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  • #2
What formula are you using?
 
  • #3
Work done is force multiplied by distance, in the case tension multiplied by extension/compression. As tension is given by:
[tex] T = \frac{kx}{l} [/tex]
Then the work done or elastic potential is given by:
[tex] \int_{0}^{e} \frac{kx}{l} \Rightarrow E_p = \frac{ke^2}{2l} [/tex]
Where e is extension and l is origional length.

I imagine you will also need to factor in gravity.
 
  • #4
Hootenanny said:
Work done is force multiplied by distance, in the case tension multiplied by extension/compression. As tension is given by:
[tex] T = \frac{kx}{l} [/tex]
Then the work done or elastic potential is given by:
[tex] \int_{0}^{e} \frac{kx}{l} \Rightarrow E_p = \frac{ke^2}{2l} [/tex]
Where e is extension and l is origional length.

I imagine you will also need to factor in gravity.

I don't know the original length, unless there's a way to determine that from the spring constant and compression.

I also can't think of how I can factor in gravity. I had no trouble with a problem where I calculated the height that a ball was shot by a vertical spring. I just calculated the work done by the spring and found the velocity from W=K2-K1. I didn't need to do anything with gravity until the ball had left the spring. Why wouldn't that be the case here aswell?
 
  • #5
You can work out the origonal length of the spring because you know the force applied to it, the weight of the block (gravity). Don't forget to resolve.
 
  • #6
Sigh, I got it. I was dividing the 59.2 N/cm by 100 instead of multiplying. Sorry about that.

As an aside, how do you make those equation images? Are you using excel, or is there something more convenient?
 
  • #7
scavok said:
https://chip.physics.purdue.edu/protected/Halliday6Mimg/h8p19.gif
(ignore values in image)

A 2.4 kg block is placed against a spring on a frictionless 17o incline. The spring, whose spring constant is 59.2 N/cm , is compressed 35.3 cm and then released.

What is the elastic potential energy of the compressed spring in J?


According to my calculations and my book, the potential energy of a spring is .5kx2. That didn't work, so I converted cm to m in k and x. That didn't work either. The only difference between this problem and a similar one in the book is that this spring is at an angle. Could someone give me a hint as to what the trick with this problem is?

The elastic spring potential energy does not take any gravitational potential energy into account by definition. So, as you said, [tex]S=(1/2)kx^2[/tex].
Thus:
[tex]k= \frac{59.2N}{1cm} * \frac{100cm}{1m} = 5920N/m [/tex]
[tex]S=(1/2)(5920N/m)(0.353m)^2=368.8J [/tex].

If that isn't the answer your book gives, your book is incorrect.

-Dan
 
  • #8
What is the answer you book gave? You can type them directly using a code called Latex. There are a few tutorials on these forums about using it. Click on equation to display the code used.
 

1. What is potential energy of a spring?

The potential energy of a spring is the energy stored in a compressed or stretched spring. It is also known as elastic potential energy, as it is the energy that the spring possesses due to its ability to return to its original shape after being deformed.

2. How is the potential energy of a spring calculated?

The potential energy of a spring can be calculated using the formula PE = 1/2kx², where k is the spring constant and x is the displacement of the spring from its equilibrium position. The spring constant is a measure of the stiffness of the spring and is typically given in units of N/m.

3. What factors affect the potential energy of a spring?

The potential energy of a spring is affected by two main factors: the spring constant and the displacement of the spring. A higher spring constant or a larger displacement will result in a greater potential energy stored in the spring.

4. What happens to the potential energy of a spring when it is compressed or stretched?

When a spring is compressed or stretched, its potential energy increases. This is because energy is stored in the spring as it is being deformed. When the spring is released, this potential energy is converted into kinetic energy, causing the spring to move back to its original shape.

5. Can the potential energy of a spring be negative?

Yes, the potential energy of a spring can be negative. This occurs when the spring is stretched beyond its equilibrium position, resulting in a negative displacement. In this case, the potential energy is also negative, indicating that the spring is being pulled in the opposite direction of its natural position.

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