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\section*{Dr René Physique 3}
\[
\tan \varphi = -\frac{V_0}{X_0 \omega_0}
\]
\[
\tan \varphi = -\frac{(-9)}{9 \sqrt{2} \times 100} = 1 \Rightarrow \varphi = \frac{\pi}{4} \text{ rad}
\]
Et l'amplitude $X_m$ est telle que
\[
\sin^2 \varphi + \cos^2 \varphi = \frac{V_0^2}{X_m^2} + \frac{X_0^2}{X_m^2} = 1
\]
\[
X_m^2 = X_0^2 + V_0^2 \quad \text{; mais } X_0^2 = V_0^2 - (L \cdot g)^2 = X_m^2 = 2X_0^2
\]
\[
X_m = \sqrt{2} \cdot X_0 \quad (m) \Rightarrow X_m = 9 \sqrt{2} \cdot 10^{-2} \text{ m}
\]
L'équation horaire du mouvement des $s$ est $x(t) = 9\sqrt{2} \cdot 10 \cos(10t + \frac{\pi}{4})$
2) Calculons à la date $t = 5s$ :
2-1) La vitesse du ootide :
\[
\text{On sait que } v(t) = -X_m \omega_0 \sin(\omega_0 t + \varphi) = -9\sqrt{2} \cdot 10 \times 10 \sin(10t + \frac{\pi}{4})
\]
\[
v(t) = -9\sqrt{2} \sin(10t + \frac{\pi}{4}) \text{ ou } v(t) = 9\sqrt{2} \cos(10t + \frac{3\pi}{4})
\]
\[
-\sin d = \cos(d + \frac{\pi}{2}), -\sin(10t + \frac{\pi}{4}) = \cos(10t + \frac{\pi}{4} + \frac{\pi}{2}) = \cos(10t + \frac{3\pi}{4})
\]
\[
\text{A } t = 5s \Rightarrow v = 9\sqrt{2} \cos(10 \times 5 + \frac{3\pi}{4}) = 9\sqrt{2} \cos(502,355^\circ) = 9\sqrt{2} \times 0,955 = 9\sqrt{2} \times 0,955
\]
\[
v = 12,2 \text{ m.s}^{-1}
\]
2-2) La position du ootide : $x = 9\sqrt{2} \cdot 10^2 \cos(10t + \frac{\pi}{4})$
\[
\text{A } t = 5s \Rightarrow x = 9\sqrt{2} \cdot 10^2 \cos(100 \times 5 + \frac{\pi}{4}) = 9\sqrt{2} \cdot 10^2 \cos(500,785^\circ)
\]
\[
|x| = -3,74 \cdot 10^{-2} \text{ m}
\]
3) Étude du mouvement du solide sur la pente ; F I
3-1) Examinons $F$ en fonction $N_F; V_I; m; g; L$ et $d$.
- méthode ; d'après le TEC entre F et I, $\Delta E_C(F \rightarrow I) = \frac{1}{2} m v_I^2 - \frac{1}{2} m v_F^2$
\[
\Delta E_C(F \rightarrow I) = \frac{1}{2} m (v_I^2 - v_F^2)
\]
\[
\leq W_F^P(F \rightarrow I) = W_F^P + W_R^P = mgL\sin d - fL \quad \text{donc}
\]
\[
\frac{1}{2} m (v_I^2 - v_F^2) = mgL\sin d - fL \Rightarrow f = m(g\sin d - \frac{v_F^2}{2L})
\]
\[
3-2) f = 0,5(10\sin 100^\circ - \frac{(5^2 - 2)}{2 \times 2}) \Rightarrow f = 15,745 \text{ N}
\]
\end{document}
