# 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

- JimLv 74 years agoFavorite 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).