# Mechanical work done when standing up

Suppose you squat. As you stand up, about how much mechanical work do your muscles do?

Your weight is $$500 N$$. Your center of mass moves at a constant speed from a height of $$H_{CM}=0.2 m$$ to $$H_{CM}^′=1.2 m$$ in $$1s$$. The answer is between $$500J$$ to $$525J$$.

$$W_{mech}=ΔK+ΔU$$ so solving for $$\Delta U=500N\cdot(1.2m-0.2m)=500J$$ is trivial. I am confused about solving for kinetic energy. At first you are stationary so $$v_0=0$$ so $$\Delta K=0.5\frac{w}{g}v^2$$. Then the explanation says:

"It is somewhat unclear what happens to your kinetic energy as you reach your full height and come to rest. Either your muscles can relax as you near your full height, using gravity to slow your motion, in which case there is a transformation of kinetic into potential energy. Alternatively, friction forces can oppose your upward motion, dissipating your kinetic energy as heat, and none of your kinetic energy is converted to potential energy.".

I think in former scenario the kinetic energy turns into potential energy at maximum height where it is stationary too so $$\Delta K=0J$$ thus mechanical energy is $$500N$$. But I don't understand why in latter scenario the $$\Delta K=25J$$. Mechanical energy just accounts for change in kinetic and change in potential energy so I think change in K is still $$0J$$ since velocity at initial and final is $$0\frac{m}{s}$$. Am I wrong? If I am wrong why am I wrong?

• "Mechanical energy just accounts for change in kinetic and change in potential energy". The problem lies here. When a rolling ball comes to a stop, where does the kinetic energy go? At the final state, there's no kinetic energy and no potential energy. It goes into thermal energy. You need another term in your equation to account for that. Commented Nov 5, 2021 at 17:33
• @Chemomechanics Show me how it is done. I know it goes to thermal equation but why 525J? That's where I am confused how do I represent mechanical energy formula with thermal energy. Commented Nov 6, 2021 at 0:55
• Just add a term for thermal energy. The work done on the system increases the sum of the kinetic, potential, and thermal energy. Now you can accommodate friction losses. Commented Nov 6, 2021 at 2:42
• @Chemomechanics So you mean $E_{mech}=\Delta K+\Delta K+Thermal$? But definition for Mechanical Energy is Change in Kinetic plus potential energy. I am confused. Commented Nov 6, 2021 at 3:21
• Sorry I misunderstood mechanical work and energy. Commented Nov 6, 2021 at 12:04