Problem — Work of a cyclist going uphill

Work for cyclists. Explain why a cyclist pedalling at constant speed uphill does positive work, and where the muscular energy ends up.

Solution

The work is positive. The cyclist pushes on the pedals to move forward: the muscular force (transmitted to the wheel) is in the same direction as the displacement, i.e. as the velocity $\vv{v}$. With angle $\alpha \approx 0$ between the driving force and the displacement,

$W_{\text{musc}} = F_{\text{musc}}\,s\cos\alpha > 0$

Where the energy goes. The speed is constant, so the kinetic energy does not change:

$\Delta E_k = 0$

The muscular energy does not accumulate as faster motion; by the energy balance it splits into two channels:

$\ev{W_{\text{musc}} = \Delta E_p + Q = mgh + Q}$

that is, part becomes gravitational potential energy $mgh$ (the cyclist rises in height $h$) and part becomes heat dissipated by friction (chain, bearings, air resistance, tyre deformation). At steady state, all the power developed by the muscles equals the sum of the power spent climbing and that dissipated: no acceleration, but real and continuous work.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Lavoro di una forza · Conservazione dell’energia meccanica Competenze: Bilancio energetico a bolle