The spring is used in a spring gun to project a 10.0 g plastic ball. In the gun, the spring is compressed by 10.0 cm before the gun is fired. How fast is the ball traveling after the gun is fired?

5.831 m/s

Explanation:

According to the work-energy law,

Work done between two points = Change in kinetic energy between the two points.

Since the plastic ball is initially at rest, its initial kinetic energy is 0 since the initial velocity = 0

Work done by the spring = ∫ F.dx

The spring is compressed by 10 cm, so, we integrate from -0.1 m to 0 m

Fₓ(x) = (-30.0 N/m)x+ (60.0 N/m²)x²

F = -30x + 60x²

W = ∫ F.dx = ∫ (-30x + 60x²) dx

W = [- 15x² + 20x³]⁰₋₀.₁ = 0 - [- 15(0.01) + 20(-0.001)] = 0.17 J

W = ΔKE

ΔKE = (mv²/2) - 0

mv²/2 = 0.17

m = 10 g = 0.01 kg

0.01 v² = 0.34

v² = 34

v = 5.831 m/s

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