How High Will a Pebble Go When Shot from a Compressed Spring?

[truthisfreedom3]

The question I'm doing for homework that I can't get is...

A spring is compressed 10.0 cm by an average force of 50.0 N. If the spring shoots a 20.0 g pebble straight up into the air, how high will it rise?

It would be sooper awesome if you could help with this!

 

[Doc Al]

Where did you get stuck? Show your work to get help. (Hint: Spring PE gets transformed to Gravitational PE.)

 

[quicknote]

I can certainly provide some insight into this problem. Conservation laws are fundamental principles in physics that state that certain properties, such as energy and momentum, remain constant in a closed system. In this case, we can use the law of conservation of energy to solve this problem.

The spring in this scenario has potential energy stored in it due to its compression. When the pebble is released, this potential energy is converted into kinetic energy as the pebble is propelled upward. According to the law of conservation of energy, the total amount of energy in the system (spring and pebble) must remain constant.

To solve for the height the pebble will reach, we can use the equation for potential energy: PE = mgh, where m is the mass of the pebble, g is the acceleration due to gravity, and h is the height. We also know that the potential energy stored in the spring is equal to the work done by the force, which is given by the equation W = Fd, where F is the average force applied and d is the distance the spring is compressed.

So, we can set up the following equation:

PE of spring = PE of pebble
F * d = mgh

Plugging in the given values, we get:

50.0 N * 0.10 m = (0.020 kg)(9.8 m/s^2)h

Solving for h, we get h = 25.5 m.

Therefore, the pebble will reach a height of 25.5 meters when shot into the air by the compressed spring. I hope this helps with your homework and further understanding of conservation laws. Keep exploring and asking questions!