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

[scavok]

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 17^o 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?

 

[Hootenanny]

What formula are you using?

 

[Hootenanny]

Work done is force multiplied by distance, in the case tension multiplied by extension/compression. As tension is given by:
$$T = \frac{kx}{l}$$
Then the work done or elastic potential is given by:
$$\int_{0}^{e} \frac{kx}{l} \Rightarrow E_p = \frac{ke^2}{2l}$$
Where e is extension and l is original length.

I imagine you will also need to factor in gravity.

 

[scavok]

Work done is force multiplied by distance, in the case tension multiplied by extension/compression. As tension is given by:
$$T = \frac{kx}{l}$$
Then the work done or elastic potential is given by:
$$\int_{0}^{e} \frac{kx}{l} \Rightarrow E_p = \frac{ke^2}{2l}$$
Where e is extension and l is original 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?

 

[Hootenanny]

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.

 

[scavok]

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?

 

[topsquark]

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 17^o 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, $$S=(1/2)kx^2$$.
Thus:
$$k= \frac{59.2N}{1cm} * \frac{100cm}{1m} = 5920N/m$$
$$S=(1/2)(5920N/m)(0.353m)^2=368.8J$$.

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

-Dan

 

[Hootenanny]

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.