Problem — Work and direction of the suitcase

Work and direction. A person carries a $20\;\mathrm{kg}$ suitcase for $50\;\mathrm{m}$ along a horizontal path, keeping it at constant height. How much work did they do against gravity? Explain why the arm gets tired even though the mechanical work is zero.

Solution

Data: $m = 20\;\mathrm{kg}$, horizontal displacement $s = 50\;\mathrm{m}$, constant height ($\Delta h = 0$).

The work of a force is $W = \vv{F}\cdot\vv{s} = F\,s\cos\alpha$, where $\alpha$ is the angle between the force and the displacement.

The weight force $\vv{F}_g$ is vertical (downward), the displacement $\vv{s}$ is horizontal: between them there is a $90^\circ$ angle, so the scalar product is zero.

$W_{\text{gravità}} = F_g\,s\cos 90^\circ = mg\,s\cdot 0 = 0$

$\ev{W_{\text{gravità}} = 0}$

Why the arm gets tired. A muscle is not a rigid hook. To stay contracted it performs continuous microscopic cycles of attachment and detachment of the actin-myosin bridges, each of which consumes ATP that degrades into heat. This is internal static tension, not macroscopic mechanical work: the point of application of the force does not move forward in the direction of the force, so $W = 0$; the metabolic energy spent is nevertheless real and is dissipated as heat. Zero mechanical work does not mean zero effort.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Lavoro di una forza Competenze: Impostazione simbolica