# How Far Does a Block Fall When Released from an Unstretched Spring?

• mathewings
In summary: The potential energy is also proportional to the square of the extension, so you can find it by multiplying the spring constant by the square of the extension.In summary, the block suspended from a spring hanging off the ceiling will oscillate between positions A and B when released. With a mass of 0.5kg and a spring constant of 10N/m, the distance the block will fall can be found by equating gravitational and spring potential energy and using conservation of energy. The potential energy is proportional to the square of the extension of the spring, so the distance can be determined by multiplying the spring constant by the square of the extension.
mathewings
A block is susoended from the end of a spring. The spring is hanging off the ceiling. An external force holds the block so that, initially, the spring is not stretched or compressed. When the block is released, it oscillates up and down between positions A and B. If the mass of the block is 0.5kg and the spring constant is 10N/m, how far will the block fall when it is released?

I've tried equating gravitational potential and spring potential energy, but the height of the block can't be determined. I don't believe there's enough information to isolate a velocity to find kinetic energy. I'm fairly certain I'm on the right track in that I need to use conservation of energy. Any suggestions would be appreciated.

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mathewings said:
A block is susoended from the end of a spring. The spring is hanging off the ceiling. An external force holds the block so that, initially, the spring is not stretched or compressed. When the block is released, it oscillates up and down between positions A and B. If the mass of the block is 0.5kg and the spring constant is 10N/m, how far will the block fall when it is released?

I've tried equating gravitational potential and spring potential energy, but the height of the block can't be determined. I don't believe there's enough information to isolate a velocity to find kinetic energy. I'm fairly certain I'm on the right track in that I need to use conservation of energy. Any suggestions would be appreciated.
You are on the right track. The spring has no PE in the initial configuration. You are free to pick the zero of gravitational potential energy anywhere you like. If you say it is zero initially, then it becomes negative as the mass falls. If you prefer, you can call it zero at the bottom and use the unknown distance it falls as the initial height. Either way, the change in potential energy is proportional to the falling distance, and the falling distance is the extension of the spring.

I would suggest using the conservation of mechanical energy principle to solve this problem. This principle states that the total mechanical energy (potential energy + kinetic energy) of a system remains constant in the absence of non-conservative forces, such as friction.

In this case, the initial mechanical energy of the system is only potential energy, as the block is not moving. As the block falls and oscillates between positions A and B, the potential energy is converted into kinetic energy and back to potential energy. Therefore, we can set the initial potential energy equal to the final potential energy at position B.

Using the equation for potential energy, PE = mgh, we can determine the initial potential energy of the system as mgh = (0.5 kg)(9.8 m/s^2)(h), where h is the initial height of the block.

At position B, the block is at its lowest point and has no potential energy, so the final potential energy is 0.

Therefore, we can set the initial potential energy equal to 0 and solve for h:

0.5 kg * 9.8 m/s^2 * h = 0

h = 0 m

This means that the block does not fall at all when it is released, as it is already at its lowest point. The oscillation between positions A and B is due to the spring force, not gravity.

In conclusion, the block will not fall when it is released, as the spring force will keep it suspended in its original position.

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