# Drop of a weight on a spring

## Homework Statement

A safe (mass = 1.00*10^3 kg) is suspended a height (d) above the top end of the spring (spring constant = 27800 N/m). The rope holding the safe breaks and the safe falls, compressing the spring a total distance of 1.80 m.
What is the initial height (d) of the safe when it was suspended above the spring?

F = ma
Fs = -kx
Fg = mg

## The Attempt at a Solution

I have only calculated Fg (9800N) and Fs (50,040N). I don't know where to go from here to get the height. Any help is appreciated, thank you.

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gneill
Mentor
Hi Abdurrr,

Welcome to Physics Forums!

Hint: Consider conservation of energy. One way to do this is with conservation of energy. Has your course explained about the energy in a compressed spring?

This is an impact situation, so it is most unlikely that energy is truly conserved. That said, conservation of energy is still a good approach, particularly if the mass of the spring is relatively small. Be sure to include the total change in gravitational potential energy, both before and after the impact, when doing the calculation.

It would be easiest to solve this using forms of energy - at the top, before the safe falls and it's hanging from a rope, the system has only Potential Energy - Gravitational (formula: m*g*h). Once it falls, after it has compressed the spring, it has Potential Energy - Spring (formula: (1/2)*k*Δs^2). K being spring constant, Δs is change in spring deformation.

Remember, it aso has gravitational energy after it has compressed the spring.