In a pinball machine, the launching spring with a spring constant of 30 N/m is compressed 0.15 m. How fast will it launch a 0.20 kg pinball?

1.84 m/s

Explanation:

Applying,

The kinetic energy of the pinball = Elastic energy of the spring

mv²/2 = ke²/2

mv² = ke² Equation 1

Where m = mass of the pin ball, v = velocity of the pin ball, k = force constant of the spring, e = extension of the spring.

make v the subject of the equation

v = √(ke²/m) Equation 2

From the question,

Given: e = 0.15 m, k = 30 N/m, m = 0.20 kg

Substitute these values into equation 2

v = √[(30×0.15²)/0.2]

v = 1.84 m/s