Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Physics, please help? Thank you, 10pts.?

A toy gun uses a spring with a force constant of 295 N/m to propel a 8.5-g steel ball. Assuming the spring is compressed 7.6 cm and friction is negligible, answer the following questions:

Randomized Variables

k = 295 N/m

m =8.51 g

d1 = 7.6 cm

Part (a) What is the magnitude of the force, in newtons, that is needed to compress the spring?

Part (b) What is the maximum height, in meters, to which the ball can be shot?

Part (c) What is the gun's maximum range, in meters, on level ground?

1 Answer

Relevance
  • ?
    Lv 7
    5 years ago
    Favorite Answer

    F = kx = 295(0.076) = 22.4 N ANS (a)

    SPE = 1/2kx² = (0.5)(295)(0.076)² = 0.85196 J

    GPE of ball = mgh =(8.5E-3)(9.81)h = 0.083385h

    by Conservation of Energy:

    GPE of ball = SPE

    0.083385h = 0.85196

    h ≈ 10.2 m ANS (b)

    The ball leaves the gun with a speed = Vexit: {KE of ball = SPE}

    Vexit = √2KE/m = √2(0.85196)/8.5E-3 = √0.200461E3 = √200.461 = 14.2 m/s

    max range of ball is obtained {when air friction is ignored} at a 45° gun angle:

    Vx = constant = horizontal ball speed component = 14.2(cos 45°) ≈ 10.0 m/s

    Voy = initial vertical ball speed {component} = 14.2(sin 45°) ≈ 10.0 m/s

    time of ball's flight = T = 2Voy/g = 20.0/9.81 ≈ 2.04 s

    max range of ball = Vx(T) = 10.0(2.04) = 20.4 m ANS (c)*

    *Comment: Note that U could just jump to ans(c) by doubling the max height, ans (b).

Still have questions? Get your answers by asking now.