# Potential energy involving both elastic and gravity

A block of mass $$m$$ is placed against an ideal spring as shown. Initially the spring (of force constant $$k$$) is compressed by a distance $$s$$. The block is then released and slides a distance d up the $$53.1^\circ$$ incline to point $$P$$. At $$P$$ the spring is relaxed and is no longer in contact with the block. There is friction between the block and the incline with coefficient $$\mu_k.$$

I am planning on teaching a course on Physics next year so I was going over a question from my final in undergrad. It appears as though I got full marks on this question, but in attempting the question again it seems that I completely missed something: the potential energy of the spring. If I were to re-do this question here is what I would do... $$K_{0}+(U_{s0}+U_{g0})+W=K_f+(U_{sf}+U_{gf})$$ $$\implies 0+(\frac{1}{2}ks^2+0)-\frac{3}{5}\mu_k mgd=\frac{1}{2}mv^2 +(0+\frac{4}{5}mgd)$$ $$\implies v=\sqrt{\frac{k}{m}s^2 -\frac{gd}{5} \left(6\mu_k+8\right)}$$ Admittedly, my solution from 5 years ago (i.e. $$v=\sqrt{\frac{dg(5-3\mu_k)}{5}}$$) looks much better which makes me suspicious that I have just lost all of my physics knowledge.

My question is this:

Is my current self correct or is my former self correct?

This type of question tends to get deemed as "illegal homework help" but I am genuinely wondering if both the potential energy done by the spring and the potential energy done by gravity must be involved here. If I was correct 5 years ago, then what allowed me to simply ignore the potential energy done by the spring?

EDIT AS PER REQUESTED: My question really boils down to the following:

If there are two types of potential energy, in this case elastic and gravitational, do we replace the $$U$$ in the conservation formula $$K_0+U_0+W=K_f+U_f$$ with $$U_0=U_{s0}+U_{g0} \text{?}$$

• People who can't access text in images have no idea what you're asking about. Please include all relevant information in the body of your question. Commented Jul 6, 2023 at 4:17
• @Chemomechanics Okay I will type it out. Commented Jul 6, 2023 at 4:18
• I think your former self is wrong. You basically computed the final speed of the block if it had been released at the top of the ramp and had slid down a distance $d$ against friction. Commented Jul 6, 2023 at 5:48
• Setup seems correct. For reasons which may be aesthetic, I would have put the $W$ term on the right side of the equation, reflecting that that amount of work has been disippated by friction in final state $f$. But it's mathematically the same Commented Jul 6, 2023 at 5:56
• No, there is no need to delete this post. Yes, it is correct. In practice, what happens is that you identify more and more potential energies and just include them in the energy balance. Commented Jul 8, 2023 at 3:51

The Energy $$~E~$$ is

$$E=\frac 12\,v^2+\frac 12\,\omega^2\,s^2- g\,(\sin(\phi)+\cos(\phi)\,\mu)\,s$$

where $$~\omega^2=\frac km~$$

at $$~t=0~$$, $$E_0=E(v=0~,s=s_0)~$$ and at point P ,$$~E_P=E(s=d)~$$

solve $$~E_P=E_0~$$ for the velocity $$~v~$$

$$\boxed{\,v_P^2=\left(s_0^2-d^2\right)\omega^2+ \left(\frac 85+\frac 65\,\mu\right)\,g\,(d-s_0)\,}$$

with $$~\sin(\phi)=\frac 45~,\cos(\phi)=\frac 35$$