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## Homework Statement

A 150 kg mass is attached to a vertical spring, stretching it by 15.0 cm. This same spring is fastened to the floor and compressed by 25.0 cm. A 5.00 kg mass is then placed on the spring. If the spring is released and the mass is launched straight up into the air by the spring, how high will the mass go above the spring's equilibrium position. Assume that the mass is released at the spring's equilibrium position.

## Homework Equations

Initial Mechanical Energy = Final Mechanical Energy

KE(i) + Gravitational PE(i) + Spring PE(i) = KE(f) + Gravitational PE(f) + Spring PE(f)

## The Attempt at a Solution

Using the first part of the experiment to find the spring constant:

m = 150 kg

xi = 0 m

xf = .15 m

g = 9.8 m/s^2

Gravitational PE (i) = Spring PE(f)

mgh = .5kx^2

[2(150 kg)(9.8 m/s^2)(.15 m)]/(.15 m)^2 = K

19600 = K

For the second part:

m = 5 kg

xi = 0 m

xf = .25 m

g = 9.8 m/s^2

Gravitational PE (i) + Spring PE (i) = Gravitational PE (f)

mgh + .5kx^2 = mgh

(5 kg)(9.8 m/s^2)(.25 m) + .5(19600)(.25)^2 = (5 kg)(9.8 m/s^2)(h final)

12.75 m = h

Does this attempt at the solution seem right?

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