La sfirsitul planului se produce o schimbare de energie cinetica si potentiala elastica, deci nu e greu de inteles ca trebuie sa aflam viteza corpului la baza planului folosind o ecuatie a acceleratiei. Rezolvarea va arata cam asa:
[tex]Se~da:\\ \\
m=4kg\\ \\
\alpha=30^\circ\\ \\
l=10m\\ \\
\mu=0,2\\ \\
k=1000\frac Nm\\ \\
g=10\frac m{s^2}\\ \\
\Delta l=?m\\ \\ \\[/tex]
[tex]Formule:\\ \\
E_c=E_p\\ \\
\frac {m\times v^2}2=\frac {k\times \Delta l^2}2\\ \\
\Delta l=v\times \sqrt \frac{m}{k}\\ \\ \\
l=\frac {v^2}{2\times a}\\ \\
v=\sqrt{2\times l\times a}\\ \\ \\
G_x-F_{fr}=m\times a\\ \\
G\times \sin\alpha-\mu\times G\times\cos\alpha=m\times a\\ \\
a=g\times(\sin\alpha-\mu\times\cos\alpha)\\ \\ \\
v=\sqrt{2\times l \times g\times(\sin\alpha-\mu\times\cos\alpha)}\\ \\
\Delta l=\sqrt{2\times l \times g\times(\sin\alpha-\mu\times\cos\alpha)\times\frac{m}{k}}\\ \\ \\[/tex]
[tex]Calcule:\\ \\
\Delta l=\sqrt{2\times 10 \times 10\times(\sin30-0,2\times\cos30)\times\frac{4}{1000}}\approx 0,51m[/tex]
