Solving a Spring-Mass System Energy Problem

[Dooga Blackrazor]

I have 3 questions. I will give you what I think is the answer and you can either give me the answer or lead me in the right direction. I appreciate both equally. With direction I can find the answer, and with the answer I can find the direction, so I am cool either way.

When the compression of a spring is reduced to half it's original value, the potential energy stored in the spring is:

1/8 original value.
1/4 original value.
The same as above.

I think it is the same because, if I remmber right, E = 1/2mv^2 and isn't affected by compression.

The total energy of a spring undergoing SHM could be expressed as:
E = 1/2mv^2, correct?

This is the one I am clueless on:

A 2.5 kg object is attached to a spring of force constant k = 4.5 kN/m. The spring is streched 10 cm from the equilibrium and released. What is the kinetic energy of the mass-spring system when the mass is 5 cm from tits equilibrium position?

14 J
11 J
17 J

No clue. I think I am using the wrong formulas or factoring in A and Xmax incorrectly or something. Any help would be great on this one.

Thanks

 

[Päällikkö]

Potential energy of a spring: (1/2)kx².
E = K + U, where K is kinetic energy and U potential energy, so E = (1/2)mv² is incorrect.

Can you now solve the problem?

 

[quicknote]

for your questions, I am happy to help.

1. The potential energy stored in a spring is directly proportional to the square of its compression or extension. Therefore, if the compression is reduced to half its original value, the potential energy will be reduced to 1/4 of its original value. So the correct answer is 1/4 original value.

2. The total energy of a spring undergoing SHM can be expressed as the sum of its kinetic energy and potential energy. Therefore, the correct expression for total energy would be E = 1/2kx^2 + 1/2mv^2, where k is the spring constant, x is the displacement from equilibrium, and v is the velocity.

3. To calculate the kinetic energy of the mass-spring system when the mass is 5 cm from its equilibrium position, we need to first calculate the velocity of the mass at that point. We can use the equation v = √(k/m)x, where k is the spring constant and m is the mass. Plugging in the given values, we get v = √(4.5 kN/m / 2.5 kg)(5 cm) = 3 m/s. Now we can use the formula for kinetic energy, K = 1/2mv^2, to get K = 1/2(2.5 kg)(3 m/s)^2 = 11.25 J. So the correct answer is 11 J.