Block on slope. Different answers using energy vs forces

[laser]

Homework Statement

(b) A stone weighing 160 N is resting on a spring at the bottom of a ramp which is inclined at θ=50 to the horizontal. The spring constant is 7000 N m^-1. How much is the spring compressed by the block?

Relevant Equations

F = ma
F = kx
E = mgh
E = 1/2kx^2

The first equation is when I use forces. The block is in static equilibrium, therefore the spring force should balance the gravitational force.

The second equation is when I use energy principles. Energy before compression = Energy at compression. The height before is x * sintheta, and the height after is 0 (defining my zero potential to be at max compression). Why do I get different answers?

 

[Orodruin]

Your energy method is wrong. If you released a block from rest where the spring elongation is zero, it would start undergoing simple harmonic motion around the equilibrium point. Your requirement that the change in gravitational potential should equal the change in potential energy of the spring is the same as requiring the mass to be at rest - which it will momentarily be at the turning point - then it will go back up. The equilibrium point that is being asked for is in the middle between the turning points - accounting exactly for the factor of two.

TLDR; Your energy method is not applicable to this case as it assumes something which is not true.

 

[Orodruin]

Another way of seeing it: You are looking for the minimum of the total potential energy, not for where the potential energy is the same as when the spring is unloaded.

 

[Chestermiller]

The difference is the kinetic energy that you get if you release it from rest, or the work you have to do to lower it gradually.

 

[erobz]

Examine a vertical spring where you simply release the mass allowing it to fall into the spring using energy conservation. When the mass fully compresses the spring it has instantaneously zero velocity like your energy equation suggest, but then ask yourself what happens next? Compare the two scenarios.

 

[kuruman]

Mechanical energy conservation works but you applied it incorrectly. You stated mechanical energy conservation as

Energy before compression = Energy at compression.

which can be misleading. A preferred way to write it is in terms of energy changes, $$\Delta K+\Delta U$$ between a starting point A and an ending point B. Here, you want to find the distance by which the spring is compressed when the mass is at equilibrium. Call that distance $x=x_{eq}$.

Suppose you release the block from rest at point A which is the relaxed position of the spring. The block will descend acquiring kinetic energy at the expense of potential energy until it reaches $x_{eq}.$ However, it will not stop there because at the equilibrium point the acceleration is zero but not the speed. At the equilibrium point the mass has acquired maximum kinetic energy $K_{max}$. It is maximum because past that point, the acceleration changes direction and the speed and kinetic energy decrease.

Now all forces are conservative. Therefore, if the mass gains kinetic energy $K_{max}$ after descending distance $x_{eq}$ when the acceleration is in the same direction, it will need to descend an additional distance $x_{eq}$ when the acceleration is reversed in order to lose $K_{max}$ and reach maximum compression at $x=x_{max}.$ Thus, $$x_{max}=2x_{eq}.$$ Now we write the energy conservation equation from point of release A to point of maximum compression B, $$\begin{align} & \Delta K+\Delta U_{\text{spring}}+\Delta U_{\text{grav.}}=0 \nonumber \\
& (0-0)+\left(\frac{1}{2}kx_{max}^2-0\right)+(0-mgx_{max}\sin\theta)=0 \nonumber \\
&\frac{1}{2}kx_{max}^2=mgx_{max}\sin\theta \implies x_{max}=\frac{2mg\sin\theta}{k}. \nonumber \\
\end{align}.$$ We have seen that $x_{max}=2x_{eq}.$ Therefore, $$x_{eq}=\frac{mg\sin\theta}{k}$$ which is what one gets using the force equilibrium argument.