Answer:
A) 24.7 m, B) 28.9 m/s²
Explanation:
Hooke's law states the force of a spring is equal to the spring constant times the change in length:
F = k ΔL
Solving for k:
k = F / ΔL
The spring constant is inversely proportional to the length:
k ∝ 1/L
Therefore:
k₁ L₁ = k₂ L₂
(F₁ / ΔL₁) L₁ = k₂ L₂
(mg / 1.30) (5.00) = k L
k = (5.00/1.30) (mg / L)
Initial energy = final energy
Initial gravitational energy = final gravitational energy + elastic energy
mgH = mgh + 1/2 k (ΔL)²
mg(H - h) = 1/2 k (ΔL)²
mg(60.0 - 10.0) = 1/2 k (ΔL)²
50mg = 1/2 k (ΔL)²
100mg = k (ΔL)²
The stuntman will fall a distance L and then an additional distance ΔL. We know this distance is equal to 60-10 = 50 m. L + ΔL = 50, so ΔL = 50 - L.
100mg = k (50 - L)²
100mg = k (2500 - 100L + L²)
100mg = (5.00/1.30) (mg / L) (2500 - 100L + L²)
26L = 2500 - 100L + L²
0 = L² - 126L + 2500
L = (126 ± √5876) / 2
L = 63 ± √1469
L ≈ 24.7 m, 101 m
Obviously L can't be more than 50.0 m, so L = 24.7 m.
As a mass on a spring, the stuntman will follow simple harmonic motion, so his maximum acceleration will be experienced at his minimum velocity, or at the very bottom.
∑F = ma
k ΔL - mg = ma
(5.00/1.30) (mg / L) ΔL - mg = ma
(5.00/1.30) (g / L) ΔL - g = a
(5.00/1.30) (9.81 / 24.7) (50.0 - 24.7) - 9.81 = a
a = 28.9 m/s²
