Comparing Spring Compression in Two Trials with Different Release Heights

In summary, students in a physics class conducted an experiment to observe the behavior of springs under different conditions. In trial 1, a 1.5 kg block was released from rest at the top of a ramp and collided with an uncompressed, massless spring. The maximum compression of the spring was recorded. In trial 2, the block was dropped from the same height directly above an identical spring. Both trials had effectively frictionless surfaces and the same mass and spring constant. It was asked whether the two springs experienced different levels of compression in bringing the block to rest. This question can be answered by assuming that the springs were compressed by the same amount after the collision and calculating the height needed to convert the potential energy of the block into energy
  • #1
phzzo
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Homework Statement


IMG_20150511_240522_446.jpg

Students in a physics class set up an experiment to examine the behavior of springs being compressed under various conditions. In the first trial of the experiment, students released a 1.5 kg block from rest at the top of a ramp. The block slid down the ramp, across a horizontal surface, and then collided with an uncompressed, massless spring. Students attached a marker to the spring so they could see how far it was compressed in bringing the block to rest. All surfaces in trial 1 were effectively frictionless.
In trial 2, the block was dropped from the same height directly above an identical spring. This spring was placed such that its bottom had the same height as the spring in trial 1 (as shown in the diagram). Again the students recorded the maximum compression of the spring in bringing the block to rest. Indicate which of the two springs experienced a greater compression in bringing the block to rest.

Homework Equations


Eg = mgh
Esp = 1/2kx2

The Attempt at a Solution


Each block will have a potential energy that is conserved as elastic energy. The blocks have the same mass and the springs have the same k. For each block and spring, mgh = 1/2kx2, but I'm not sure about the h for Trial 2. Would it be the same as Trial 1?
 
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  • #2
phzzo said:
I'm not sure about the h for Trial 2. Would it be the same as Trial 1?

Try assuming that in both cases the spring is compressed some amount (say 50%?) after the collision. Draw that and work out the height that you would use to calculate the PE that has been converted into energy stored in the spring. Can they be the same?
 

1. What is spring compression?

Spring compression is the process of applying a force to a spring, causing it to shorten and store potential energy.

2. How does spring compression relate to energy?

When a spring is compressed, potential energy is stored in the spring due to the work done by the applied force. This potential energy can later be converted into kinetic energy when the spring is released.

3. What factors affect the amount of energy stored in a compressed spring?

The amount of energy stored in a compressed spring is affected by the force applied, the distance the spring is compressed, and the stiffness of the spring. The formula for calculating the potential energy stored in a spring is 1/2*k*x^2, where k is the spring constant and x is the distance the spring is compressed.

4. How is energy released from a compressed spring?

Energy is released from a compressed spring when the force applied to the spring is removed. The stored potential energy in the spring is converted into kinetic energy as the spring returns to its original shape.

5. What are some real-life applications of spring compression?

Spring compression is used in various everyday objects, such as door hinges, car suspensions, and trampolines. It is also used in more advanced technologies, including shock absorbers, mechanical watches, and robotic limbs.

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