Problem — Cart on a smooth track

Problem

A $50\;\mathrm{kg}$ cart starts from rest at the top of a track $5{,}0\;\mathrm{m}$ high and descends along a smooth surface. What is its speed at the bottom?

Solution

Smooth track: mechanical energy is conserved. The initial gravitational potential energy all becomes kinetic energy at the bottom: $m\,g\,h = \frac{1}{2}\,m\,v^2$ The mass cancels out (the result does not depend on the $50\;\mathrm{kg}$): $v = \sqrt{2\,g\,h} = \sqrt{2\cdot 9{,}8\;\mathrm{m/s^2}\cdot 5{,}0\;\mathrm{m}} = \sqrt{98}\;\mathrm{m/s}$ $\ev{v \approx 9{,}9\;\mathrm{m/s}}$

Collegamenti

Argomenti: Lavoro ed energia Concetti: Conservazione dell’energia meccanica Competenze: Conservazione dell’energia Oggetti: Blocco