Calculating Initial Height of a Falling Safe on a Spring

In summary, the conversation discusses a problem involving a safe falling and compressing a spring, and the question of finding the initial height of the safe before the fall. The solution involves using conservation of energy, specifically considering the change in gravitational potential energy and spring potential energy before and after the impact. The formula for gravitational potential energy is m*g*h, while the formula for spring potential energy is (1/2)*k*Δs^2, where k is the spring constant and Δs is the change in spring deformation.
  • #1
Abdurrr
1
0

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?

Homework Equations


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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  • #2
Hi Abdurrr,

Welcome to Physics Forums!

Hint: Consider conservation of energy. :wink:
 
  • #3
One way to do this is with conservation of energy. Has your course explained about the energy in a compressed spring?
 
  • #4
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.
 
  • #5
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.
 
  • #6
Remember, it aso has gravitational energy after it has compressed the spring.
 

1. What is the concept behind a drop of a weight on a spring?

The concept behind a drop of a weight on a spring involves the principles of potential and kinetic energy. When a weight is dropped onto a spring, it compresses the spring and stores potential energy. As the spring expands back to its original position, the potential energy is converted into kinetic energy, causing the weight to bounce up and down.

2. How does the height of the drop affect the motion of the weight on the spring?

The height of the drop affects the motion of the weight on the spring because it determines the amount of potential energy that is transferred to the spring. The higher the drop, the more potential energy is stored in the spring, resulting in a higher bounce and longer period of motion.

3. What factors affect the frequency of the motion of the weight on the spring?

The frequency of the motion of the weight on the spring is affected by the mass of the weight, the stiffness of the spring, and the gravitational force. A heavier weight, stiffer spring, and stronger gravitational force will result in a higher frequency of motion.

4. How does damping affect the motion of the weight on the spring?

Damping, or the dissipation of energy, affects the motion of the weight on the spring by reducing the amplitude of the oscillations. This means that over time, the weight will bounce with smaller and smaller movements until it eventually comes to a rest.

5. Can the drop of a weight on a spring be used to measure the strength of gravity?

Yes, the drop of a weight on a spring can be used to measure the strength of gravity. By measuring the period of the motion and using the equation for the period of a spring-mass system, the acceleration due to gravity can be calculated. This method is known as the "spring scale" method and is commonly used in physics experiments.

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