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?
Best Answer
Unlike the forces of nature, the force generated by muscle requires energy input. This is because muscle contraction is caused by myosin heads repeatedly binding to actin and advancing along the actin molecule. Biological processes like these are nowhere near 100% efficient: the myosin heads occasionally slip backward.
The transfer of energy from ATP to the myosin head during the power stroke is inefficient. This energy is dissipated mostly as heat, which is why animals sweat during physical activity.