Solution :
\( W = \vec{F} \cdot \vec{d} = F \times d \times \cos(0°) = 50 \times 10 \times 1 = 500 \text{ J} \)

Solution :
\( W = F \times d \times \cos(30°) = 100 \times 5 \times \frac{\sqrt{3}}{2} ≈ 433 \text{ J} \)

Solution :
\( W = F \times d \times \cos(180°) = 20 \times 15 \times (-1) = -300 \text{ J} \)
Le travail est résistant (négatif)

Solution :
\( W = m \times g \times h = 2 \times 9.81 \times 10 = 196.2 \text{ J} \)
Le travail est moteur (positif)

Solution :
\( P = \frac{W}{t} = \frac{1200}{30} = 40 \text{ W} \)

Solution :
\( P = \vec{F} \cdot \vec{v} = F \times v \times \cos(0°) = 150 \times 4 = 600 \text{ W} \)

Solution :
\( W = \int_{1}^{3} 3x^2 dx = [x^3]_{1}^{3} = 27 – 1 = 26 \text{ J} \)

Solution :
\( \eta = \frac{\text{Énergie utile}}{\text{Énergie consommée}} = \frac{600}{800} = 0.75 \) soit 75%

Solution :
\( W = \Delta E_c = \frac{1}{2}m(v_2^2 – v_1^2) = \frac{1}{2} \times 5 \times (36 – 4) = 80 \text{ J} \)

Solution :
1. \( v = 18 \text{ km/h} = 5 \text{ m/s} \)
\( F = mg \sin(\theta) ≈ 80 \times 9.81 \times 0.05 = 39.24 \text{ N} \)
\( P = F \times v = 39.24 \times 5 = 196.2 \text{ W} \)
2. \( P_{\text{fournie}} = \frac{P_{\text{utile}}}{\eta} = \frac{196.2}{0.25} = 784.8 \text{ W} \)