Respuesta :
Answer:
a) F = 21.7 N , b) h = 8.84 m , c) θ = 58.5º d) R = 15.75 m
Explanation:
a) Hook's law is
x = 7.6 cm (1m / 100cm) = 0.076 m
m = 9.5 g (1 kg / 1000g) = 9.5 10-3 kg
F = K x
F = 285 0.076
F = 21.7 N
b) We use energy conservation
Eo = Ke = ½ kx²
Ef = U = mg h
Eo = Ef
½ k x² = mg h
h = ½ k x² / mg
h = ½ 285 0.076² / 9.5 10⁻³ 9.8
h = 8.84 m
c) Let's calculate the speed with which it leaves the gun
Ke = K
½ k x² = ½ m v²
v = √ (k/m x²)
v = √( 285 / 9.5 10-3 0.076²)
v = 13.16 m / s
v₀ = v = 13.16 m/s
As we have the horizontal distance we can calculate the travel time
x = vox t
t = vox / x
t = vo cos θ / x
t = 13.16 cos θ / 3
t = 4.39 cos θ
y = [tex]v_{oy}[/tex]t - ½ g t²
0 = vo sin θ t-1/2 9.8 t²
0 = 13.16 sin θ t-4.9t²
0 = 13.16 sin θ (4.39 cos θ) - 4.9 (4.39 cos θ)²
0 = 57.78 sin θ cos θ - 94.43 cos² θ
0 = 57.78 sin θ - 94.43 cos θ
tan θ = 94.43 / 57.78
θ = 58.5º
d) The range maximum is
R = vo² sin 2θ / g
R = 13.16² sin 2 58.5 /9.8
R = 15.75 m
