Atoy 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?

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 = 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