Solution :
\( E_{pp} = m \times g \times h = 2 \times 9,81 \times 1,5 = 29,43 \text{ J} \)

Solution :
\( \Delta E_{pp} = m \times g \times \Delta h = 70 \times 9,81 \times 300 = 206\,010 \text{ J} \)

Solution :
\( W(\vec{P}) = m \times g \times h = 5 \times 9,81 \times 10 = 490,5 \text{ J} \)
Le travail est moteur (positif) lors d’une descente.

Solution :
\( E_{m1} = E_{m2} \) ⇒ \( mgh = \frac{1}{2}mv^2 \)
\( v = \sqrt{2gh} = \sqrt{2 \times 9,81 \times 4} ≈ 8,86 \text{ m/s} \)

Solution :
\( h = 100 \times \sin(20°) ≈ 34,2 \text{ m} \)
\( W_{frot} = -\mu mg \cos(20°) \times d ≈ -0,1 \times 80 \times 9,81 \times 0,94 \times 100 ≈ -7\,376 \text{ J} \)
\( \Delta E_c = W_P + W_{frot} \) ⇒ \( \frac{1}{2}mv^2 – \frac{1}{2}mv_0^2 = mgh + W_{frot} \)
\( v ≈ 22,4 \text{ m/s} \)

Solution :
\( h = L(1 – \cosθ) ≈ 0,268 \text{ m} \)
\( mgh = \frac{1}{2}mv^2 \) ⇒ \( v = \sqrt{2gh} ≈ 2,29 \text{ m/s} \)

Solution :
\( \frac{1}{2}mv^2 = mgh_{max} \) ⇒ \( h_{max} = \frac{v^2}{2g} = \frac{225}{19,62} ≈ 11,47 \text{ m} \)

Solution :
\( \frac{1}{2}kx^2 = mgh \) ⇒ \( h = \frac{kx^2}{2mg} = \frac{200 \times 0,01}{2 \times 0,5 \times 9,81} ≈ 0,204 \text{ m} \)

Solution :
Condition : \( v_{haut} ≥ \sqrt{gR} ≈ 4,43 \text{ m/s} \)
Conservation énergie : \( \frac{1}{2}mv_{bas}^2 = \frac{1}{2}mv_{haut}^2 + mg(2R) \)
\( v_{bas} = \sqrt{5gR} ≈ 9,90 \text{ m/s} \)

Solution :
1. \( W_{frot} = -\mu mg \cosθ \times L \) (cosθ = adjacent/hypoténuse)
\( \Delta E_c = W_P + W_{frot} \) ⇒ \( v ≈ 17,1 \text{ m/s} \)
2. Sans frottement : \( v = \sqrt{2gh} ≈ 19,8 \text{ m/s} \)