How do I calculate elastic potential energy?

AI Thread Summary
To calculate the elastic potential energy stored in a spring, the formula U = (1/2)kx^2 is used, where k is the spring constant and x is the stretch length. The problem involves two carts, with the 0.29-kg cart moving at 1.1 m/s after release. By applying conservation of energy and momentum, the kinetic energies of both carts are calculated, leading to a total kinetic energy of 0.44326 J. The elastic potential energy is determined to be equal to this change in kinetic energy, confirming the calculations are correct. Rounding the final answer to match significant digits is recommended for accuracy.
alicia12131415
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Homework Statement
A 0.29-kg cart and a 0.19-kg cart are held together with a compressed spring between them. When they are released, the 0.29-kg cart moves at 1.1 m/s to the right.
How much elastic potential energy was stored in the spring before the release?
Relevant Equations
U = (1/2)kx^2
k = spring force constant
x = stretch length
Is this the correct formula to use, if not what is. I really just don't know where to start. I really just need to know where and how to start.
 
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Hello @alicia12131415,

Welcome to PF! :welcome:

alicia12131415 said:
Problem Statement
A 0.29-kg cart and a 0.19-kg cart are held together with a compressed spring between them. When they are released, the 0.29-kg cart moves at 1.1 m/s to the right.
How much elastic potential energy was stored in the spring before the release?
Relevant Equations
U = (1/2)kx^2
k = spring force constant
x = stretch length

Is this the correct formula to use, if not what is. I really just don't know where to start. I really just need to know where and how to start.
Try using conservation of energy.

What's the kinetic energy of the 0.29 kg kart? What's the kinetic energy of the 0.19 kg kart? (Hint: that last calculation is also going to involve conservation of momentum.)

I'm guessing that the problem wants you to treat the spring itself as ideal and massless.
 
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So, I got

Velocity for 0.19 kg cart:
0 = (0.29kg)(1.1m/s) + (0.19)v2f
v2f = −1.6789 m/s

Kinetic energy:
0.29 kg KE1f = (1/2)(0.29kg)(1.1m/s)2 = 0.17545 J
0.19 kg KE2f = (1/2)(0.19kg)(-1.6789m/s)2 = 0.26781 J
KEf = 0.44326 J
ΔKE = 0.44326 J - 0 J = 0.44326 J

Conservation of Energy:
KEi + PEi = KEf + PEf
PEi = ΔKE + PEf
PEi = 0.44326 J + 0 J = 0.44326 J

Is this correct?
 
alicia12131415 said:
So, I got

Velocity for 0.19 kg cart:
0 = (0.29kg)(1.1m/s) + (0.19)v2f
v2f = −1.6789 m/s

Kinetic energy:
0.29 kg KE1f = (1/2)(0.29kg)(1.1m/s)2 = 0.17545 J
0.19 kg KE2f = (1/2)(0.19kg)(-1.6789m/s)2 = 0.26781 J
KEf = 0.44326 J
ΔKE = 0.44326 J - 0 J = 0.44326 J

Conservation of Energy:
KEi + PEi = KEf + PEf
PEi = ΔKE + PEf
PEi = 0.44326 J + 0 J = 0.44326 J

Is this correct?
Looks good, but you should round the answer to match the fewest significant digits of the given data.
 
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I will, thank you for checking my work. Have a wonderful weekend.
 
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