Spring Constant Question, AAMC Physics Assessment #51

[MedGrl@2022]

A vertically oriented spring is stretched by 0.15 m when a 100 g mass is suspended from it. What is the approximate spring constant of the spring?
A) 0.015 N/m
B) 0.15 N/m
C) 1.5 N/m
D) 6.5 N/m

I understand how they got to answer D from the equations F=mg=kx. Orginally, when I tried to solve this I tried to use the energy equations (1/2)kx^2=mgh…. I used 0.15m for x and h. When I solved it this way, k=13.3… why is using the energy equations for this problem incorrect?

Thank you for all your help!

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[milski]

You cannot solve this from energetic point of view, at least not in a straight forward way. What you are saying in your energy equation is that the energy of the system with the spring in its non-extend state and the weight at rest (0 potential energy in the spring and mgh in the weight) is the same as the energy of the system when spring is extended and the weight at rest (0.5kx² for the spring and 0 for the weight). That cannot really be true, or the system would not move from one state to the other. If you started with the non-extended spring, you'll have the system oscillate indefinitely (barring any friction) around the equilibrium point.

 

[milski]

Actually, using energies is not that bad but still rather convoluted.

Let's start at rest, with the spring collapsed at its original length, weight just attached to it. The total energy in this case is mgx - only potential energy for the weight is non-zero. Since the spring does not exert any force on the weight, it will start falling down, extending the spring and accelerating.

When it reaches the equilibrium point, 0.15 down, the force from the spring will be equal to the weight, the acceleration will change sign through zero and the weight will start slowing down. At the equilibrium point, the energy of the spring will be 0.5kx², the potential energy of the weight will be 0 but it will have some kinetic energy. It can be derived that this kinetic energy is also 0.5kx² - it's the potential energy of the spring being converted back and forth to KE.

The total energy of the system as the weight crosses the equilibrium point is 0.5kx²+0.5kx²+0=kx².

If you solve kx²=mgx for x=0.15 and m=0.1, you'll get k=6.5.

So it is doable but it is more of a proof that physics laws are not broken than a practical approach.

Btw, you'll notice that the energies solution effectively got us the same equation at the end - kx=mg.

 

[MedGrl@2022]

You cannot solve this from energetic point of view, at least not in a straight forward way. What you are saying in your energy equation is that the energy of the system with the spring in its non-extend state and the weight at rest (0 potential energy in the spring and mgh in the weight) is the same as the energy of the system when spring is extended and the weight at rest (0.5kx² for the spring and 0 for the weight). That cannot really be true, or the system would not move from one state to the other. If you started with the non-extended spring, you'll have the system oscillate indefinitely (barring any friction) around the equilibrium point.

so basically you are saying the .5kx^2 does not equal mgh because the spring is extended? Is that right?

mgh is when the spring is at its equilibrium point. .5kx^2 is the kinetic energy when the spring is at one extreme. In order to get the total energy of the system, you would have to multiply .5kx^2 by 2 because there are two extremes of the spring? Therefore .5kx^2 does not equal mgh but, kx^2=mgh.

Is that correct? 🙂

 

[milski]

so basically you are saying the .5kx^2 does not equal mgh because the spring is extended? Is that right?

mgh is when the spring is at its equilibrium point. .5kx^2 is the kinetic energy when the spring is at one extreme. In order to get the total energy of the system, you would have to multiply .5kx^2 by 2 because there are two extremes of the spring? Therefore .5kx^2 does not equal mgh but, kx^2=mgh.

Is that correct? 🙂

Sort of. Just to clear up the terminology - there are two equilibrium points, without weight and with weight attached. Lets call them E1 and E2.

If the system is at rest in E1 and you attach the weight, it will not be at rest when the weight reaches E2. This is why mgh is not equal to half mx².

At E2 you have half mx² potential energy from the spring and half mx² kinetic energy from the weight moving. The kinetic energy is the same as the potential energy because now you have a spring oscillating around E2 with amplitude x.

It's messy, I'll be the first one to admit it. Using energies is the easier approach when you know the initial and final velocities in a closed system. Here you need to account for stopping the spring at E2, so it's not really a good situation to use energies.