# Hooke's law vs. elastic potential energy

I am currently learning about elastic potential energy and this is a question that was given to us by my teacher:

When a 13.2-kg mass is placed on top of a vertical spring, the spring compresses 5.93 cm. Find the force constant of the spring.

To solve this question he used total mechanical energy (ME= GPE+ EPE+ KE)and the conversion of the energy from GPE to EPE (as the object falls down it's energy changes from GPE to EPE and the KE is 0 because the final and initial velocities are zero.)

Diagram of the problem

Because this is a conserved system then ME =0 deltaGPE=-deltaEPE then he substituted the given into the equations: (13.2)(10) (0-0.0593)=-1/2 (k) (0.0593^2-0^2) k= 4451.9 N/m

However by using Hooke's law (which I got from the internet- In class we still didn't take hooke's law and all what we know about springs...(elastic) is from EPE equation.): F=kx the force here is Fg=(13.2)(10)=132 132=k(0.0593) k=2225.96 N/m

So my question is why are the answers diffrent what is missing in either of these ways??

• To begin with, in your 2nd attempt 132/0.0593=2225.9, whereas you dropped an order of magnitude. From there, you're trying to explain the factor of 1/2. Feb 22, 2015 at 13:36
• Yes i fixed it now... and yes the first k is *2 the second one because of the 1/2 in the EPE formula. Feb 22, 2015 at 13:44

## 3 Answers

I believe we must trash the energy approach. The fundamental assumption in this question is that the forces will balance at the same point that the energy balances. This isn't necessarily true.

I would call upon a mental model of a mass displaced on a spring. It starts at rest, but not at the neutral-force position. What happens? It oscillates. The same should be predicted for a mass resting on top of a spring. The fact that gravitational and elastic energies don't balance reflects the fact that it won't quickly come to rest there. If there is no friction, it will go past the balanced-force point and oscillate.

To solve this question he used total mechanical energy (ME= GPE+ EPE+ KE)and the conversion of the energy from GPE to EPE (as the object falls down it's energy changes from GPE to EPE and the KE is 0 because the final and initial velocities are zero.)

You can't do this. In a proper accounting, the mass would have some kinetic energy by the point that it reaches 0.0593 cm. It will blow past that point due to inertia. It will reverse its direction at 0.1186 m. At that point, there truly is no kinetic energy left and the elastic potential balances the gravitational potential. But this is not the same point at which it will come to rest.

If you play out the oscillation for a long time, then energy is eventually dissipated and it will come to rest at 0.0593 m. By that point, half of the original gravitational energy has been converted into heat.

• good answer, but no need to discard the energy approach. just note that the spring extension when KE=0 is actually 2*0.0593 (already explained that when extension is 0.0593, the mass still has KE that was wrongly assumed to be zero). in other words the x in the energy equation (oscillating system) is different from the x in the force equation (equilibrium system) Feb 22, 2015 at 14:14
• why is KE 2*0.0593? Feb 22, 2015 at 14:39

The answer is simple. This is a case of elastic potential energy. Just apply Hooke's Law.

$$F= kx$$

where $F$ is the force due to the spring (N), $x$ is the elongation (m) and $k$ is the spring constant (N/m).

For this particular case, $F$ is the weight of the object placed on top of the spring.

$F = mg = (13.2 kg)(9.80665 m/²) = 129.45 N$

$x = 5.93 cm = 0.0593 m$

Isolating the spring constant $k$ we have:

$k = F/x = 129.45 N/0.0593 m = 2,183 N/m = 2.2 kN/m$

The answer $$k=\frac{F}{x}$$ by applying mechanical equilibrium seems to be correct. However, at the maximum compression the velocity should be zero in any case, even if the system is oscillating. Therefore, we need to calrify why the energetic approach seems to leads to a differents result, actually due to the $$\frac{1}{2}$$ factor.

• Yes, perhaps you could expand your answer to say why the energy approach leads to the different result May 3, 2021 at 17:54