[Physics] $\Delta K=\Delta U$ vs $\Delta K = -\Delta U$

conventionsenergyenergy-conservationnewtonian-mechanicspotential energy

While doing physics homework, I noticed that in some problems, the change in kinetic energy(KE) is equal to the change in potential energy(PE), even though I learned that conservation of energy shows that the change in K.E. is equal to the negative of the change in P.E. My question is, when are they equal and when are they equal to the negative of the other?

For example, a problem with a 2 block system in which one block is on a frictionless table and the other block is attached to the first block by a rope, hanging off the table. The problem gives me both masses and asks for the speed of the blocks after moving 2 meters. In the solution, it states "The only force is gravity (a conservative force), so the change in potential energy of the system is equal to the change in total kinetic energy ΔU=ΔK."

However, in another problem that involves pushing a block against a spring attached to the wall, mass, compression distance, and spring constant are given. The solution states

"ΔU = 0 – .5k(x^2) – 0 = -.5k(x^2) and that energy conservation implies ΔK=−ΔU."

Best Answer

First of all, note that the total energy of a mechanical system is always conserved for this system, i.e., the change in the total energy should always be zero: $$\Delta E=\Delta K+\Delta U=0$$ So, the correct expression is $\Delta K=-\Delta U$ and when only magnitudes are considered, we have $|\Delta K|=|\Delta U|$.

Physical Interpretation:

Consider a pulley system with two unequal masses hanging on it, a larger mass $M$ and a smaller mass $m$. Let the system is released from rest when both the masses are at the same level.

Now, the larger mass $M$ comes down and the smaller mass $m$ goes up and let their common velocity be $V$ and their heights be $H_M$ and $H_m$ respectively.

  • Total energy for mass $M$: $E_M=\frac{1}{2}MV^2+MgH_M$
  • Total energy for mass $m$: $E_m=\frac{1}{2}mV^2+mgH_m$

Since, $M>m$ and $H_M<H_m$, we have the following conclusions:

For the mass $M$, the kinetic energy $K$ increases while the potential energy $U$ decreases as the mass comes down, i.e., $\Delta K>0$ and $\Delta U<0$.

For the mass $m$, the kinetic energy $K$ decreases while the potential energy $U$ increases as the mass goes up, i.e., $\Delta K<0$ and $\Delta U>0$.

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