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3 years ago

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A toy gun uses a spring with a force constant of 285 N/m to propel a 9.5-g steel ball. Assuming the spring is compressed 7.6 cm

and friction is negligible, answer the following questions: (a) How much force is needed to compress the spring? (b) To what maximum height can the ball be shot? (c) At what angles above the horizontal may a child aim to hit a target 3.00 m away at the same height as the gun? (d) What is the gun's maximum range on level ground?

1 answer:

Bas_tet [7]3 years ago

6 0

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 = $v_{oy}$ 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

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