Solution :
Conversion : 90 km/h = 25 m/s
\( E_c = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 375\,000 \text{ J} \)

Solution :
\( W = \Delta E_c = \frac{1}{2}m(v_2^2 – v_1^2) \)
\( = \frac{1}{2} \times 5 \times (400 – 100) = 750 \text{ J} \)

Solution :
\( W = \Delta E_c = 0 – \frac{1}{2}mv^2 \) (vitesse initiale inconnue mais résultat en fonction de v)
Ou si vitesse initiale = 72 km/h (20 m/s) :
\( W = -\frac{1}{2} \times 1000 \times 20^2 = -200\,000 \text{ J} \)

Solution :
\( W = F \times d = 200 \times 20 = 4000 \text{ J} \)
\( W = \Delta E_c \Rightarrow 4000 = \frac{1}{2} \times 10 \times v^2 \)
\( v = \sqrt{800} ≈ 28,28 \text{ m/s} \)

Solution :
Hauteur : \( h = 100 \times \sin(30°) = 50 \text{ m} \)
\( mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} ≈ 31,30 \text{ m/s} \)

Solution :
\( F_f = μmg = 0,3 \times 2 \times 9,81 ≈ 5,886 \text{ N} \)
\( W = -F_f \times d = \Delta E_c = -\frac{1}{2}mv^2 \)
\( d = \frac{mv^2}{2F_f} ≈ \frac{2 \times 100}{2 \times 5,886} ≈ 16,99 \text{ m} \)

Solution :
100 km/h = 27,78 m/s
\( \Delta E_c = \frac{1}{2} \times 1200 \times 27,78^2 ≈ 463\,000 \text{ J} \)
\( P = \frac{W}{t} = \frac{463\,000}{10} = 46,3 \text{ kW} \)

Solution :
1. Départ : \( E_c = \frac{1}{2} \times 0,05 \times 20^2 = 10 \text{ J} \)
2. À mi-hauteur (h_max = v²/2g = 20,39 m) :
\( E_c = E_{c0} – mgh/2 = 10 – 0,05 \times 9,81 \times 10,195 ≈ 5 \text{ J} \)

Solution :
\( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \)
\( v = \sqrt{\frac{kx^2}{m}} = \sqrt{\frac{200 \times 0,01}{0,5}} = 2 \text{ m/s} \)

Solution :
1. \( v = 5 \text{ m/s} \), \( F = mg \sinθ ≈ 80 \times 9,81 \times 0,05 ≈ 39,24 \text{ N} \)
\( P = F \times v ≈ 196,2 \text{ W} \)
2. \( P_{\text{fournie}} = \frac{P_{\text{utile}}}{\text{rendement}} = \frac{196,2}{0,25} ≈ 784,8 \text{ W} \)