Potential Energy of a Spring

In summary: When you add an identical mass, the total force acting on the string will double, and according to Hooke's law, the displacement of the string will also double. This is because the spring constant (k) remains the same for the same spring regardless of the mass attached to it.
  • #1
alemon
3
0
New member here...

I have the following homework problem:
"A 200-g mass attached to the end of a spring causes it to stretch 5.0 cm. If another 200-gmass is added to the spring, the potential energy of the spring will be"

I know the answer is 4 times as much but I am not quite sure how the answer is being reached.

I thought I could use the forumla:
U= 1/2 kx^2

But I am wrong.

Any help would be much appreciated!
Thank you!
A.
 
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  • #2
You have the correct formula for the potential of a spring. Are you not supposed to find the final potential of the spring once the other mass is added. Tell us exactly what you've tried.

P.S. Welcome to the forums.
 
  • #3
Maybe, I did this wrong, but...

I doubled up the two 200s...

Then I took the spring constant:
K=400/5
K=20

Then I used the potential energy formula:
U=1/2(20)(5)^2
and I came up with 250

Which is probably ALL wrong.
 
  • #4
Your calculation of the spring constant is wrong. What you need to do is use the original data for 200 grams to get the spring constant. The force exerted on the spring by the mass is given by its weight (i.e. weight = mg). And then you should use the equation that relates the force exerted on a spring to the distance moved (F=-kx). Remember to use SI units aswell.
 
  • #5
Could you...

You first need to find the Spring Constant:
K=f/x=200/5=40

So your spring constant (k) is 40, your displacement (x) is the 5cm.

Now, first find your potential energy without the extra 200 grams added…
u=1/2kx^2=1/2(40)5^2= 500

Now, find your second potential energy with the added 200 grams added… Since you’ve doubled the weight (k will remain the same) BUT your displacement will increase to 10 (as you doubled).
u= kx =1/2(40)10^2 = 2,000

So, now… 2000 divided by 500 equals “4 times as much”.
 
  • #6
Well it is going to be 4 times as much but what you're doing is wrong. You can't say k =200/5 because 200 is a mass not a force. The force you want to use is the weight of the 200g mass. Weight as I said above is the mass multiplied by the gravitational acceleration (9.8 m/s2).

What the question is asking is what the potential energy will be so you want the actual value. Also can I stress again please use SI units its very much safer. So instead of grams convert to Kilograms and instead of centimeters use meters.

Other than that your method is correct.
 
  • #7
alemon,
Why don't you leave the numbers and just play with the symbols? Its much much easier and certainly not confusing at all.
You don't get confused with the SI units as well ..
Try working that way. Only subsitute the numbers at the last formula.
 
  • #8
Kurdt said:
Well it is going to be 4 times as much but what you're doing is wrong. You can't say k =200/5 because 200 is a mass not a force. The force you want to use is the weight of the 200g mass. Weight as I said above is the mass multiplied by the gravitational acceleration (9.8 m/s2).

What the question is asking is what the potential energy will be so you want the actual value. Also can I stress again please use SI units its very much safer. So instead of grams convert to Kilograms and instead of centimeters use meters.

Other than that your method is correct.

why do we assume that the string's displacement will double?
 
  • #9
robvba said:
why do we assume that the string's displacement will double?
You don't assume it, it's a consequence of Hooke's law.
 

1. What is potential energy of a spring?

The potential energy of a spring is the energy stored in a compressed or stretched spring due to its position or shape.

2. How is potential energy related to a spring?

Potential energy is directly proportional to the amount of deformation or displacement of a spring. The more a spring is compressed or stretched, the more potential energy it has.

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

The potential energy of a spring is affected by its spring constant and displacement. The higher the spring constant, the more potential energy a spring can store. The greater the displacement, the more potential energy a spring can have.

4. How is potential energy of a spring different from kinetic energy?

Potential energy is the energy stored in an object due to its position or shape, while kinetic energy is the energy an object has due to its motion. In the case of a spring, potential energy is stored when it is compressed or stretched, while kinetic energy is generated when the spring is released and returns to its original shape.

5. How is potential energy of a spring used in everyday life?

Potential energy of a spring is used in many everyday objects, such as door hinges, trampolines, and shock absorbers. It is also used in more advanced technologies, such as in mechanical watches and energy storage systems.

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