Question

Ma ajuta cineva cu problema asta ?

Answer 1

Răspuns:

a) 0,2

Explicație:

$\text{La inceput, cand sania coboara pe partie:}\\ \\ G = mg\\ \\ G_t = mg\sin{\alpha}\\ \\ G_n = mg\cos{\alpha} \\ \\ N = G_n = mg\cos{\alpha}\\ \\ F_f = \mu N = \mu mg\cos{\alpha}\\ \\ ma = G_t - F_f = mg\sin{\alpha} - \mu mg\cos{\alpha}\\ \\ a = \frac{mg(\sin{\alpha} - \mu \cos{\alpha})}{m} = g(\sin{\alpha} - \mu\cos{\alpha})\\ \\ \text{Viteza pe care o are cand ajunge la baza partiei:}\\ \\ v^2 = v_0^2 + 2ad = 0 + 2g(\sin{\alpha} - \mu\cos{\alpha}) \cdot d\\ \\ h = d\sin{\alpha} \implies d = \frac{h}{\sin{\alpha}}\\ \\ v^2 = 2g(\sin{\alpha} - \mu\cos{\alpha}) \cdot \frac{h}{\sin \alpha} \\ \\ = 2g\frac{h}{\sin \alpha}\cdot \sin \alpha - 2g\frac{h}{\sin \alpha} \cdot \mu \cos{\alpha} \\ \\ = 2gh - 2gh\mu ctg(\alpha) \\ \\ = 2gh(1 - \mu ctg(\alpha))$

$\text{Proiectia pe orizontala a partiei:}\\ \\ l = d\cos \alpha = h \cdot ctg(\alpha)\\ \\ \text{Pentru traseul pe orizontala:}\\ \\ 0^2 = v_0^2 + 2ad \\ \\ v_0^2 = 2gh(1 - \mu ctg(\alpha))$

$\text{Acum pentru acceleratia pe orizontala:}\\ \\ a = \frac{-F_f}{m} = \frac{-\mu N}{m} = \frac{-\mu G}{m} = \frac{-\mu mg}{m} = -\mu g\\ \\ 0 = 2gh(1-\mu ctg(\alpha)) - 2\mu g (d - l) \\ \\ 2\mu (d-l) = 2h(1 - \mu ctg(\alpha))\\ \\ \mu (d - h ctg(\alpha)) = h - h\mu ctg(\alpha)\\ \\ d\mu - h\mu ctg(\alpha) = h - h\mu ctg(\alpha)\Bigg|+h\mu ctg(\alpha)\\ \\ d\mu = h\\ \\ \boxed{\mu = \frac{h}{d} = \frac{10}{50} = \frac{1}{5} = 0,2}$