# Energy conservation, and mgh

• rsala
In summary, When using energy conservation, the potential energy of gravity is negative when the ball is below the set point of potential energy at 0. The initial potential energy can be calculated by adding mgh and 0.5kx^2. The total potential energy at the compressed initial part is -mgh-0.5kx^2. The potential energy of the spring is always positive at +kx^2/2. The correct formula for finding velocity once the ball leaves the gun is \sqrt{\frac{2(mgh+\frac{1}{2}kx^{2})}{m}} where x=-0.25.

#### rsala

my question is, when using energy conservation when do i put the potential energy of gravity negative? if its under where it set my potential energy to 0? or am i wrong, is gravity's potential energy always positive (except at my set u=0 point) ?

can someone explain this to me, having trouble getting ANYthing done because i do not understand how to set up these homework questionsfor example, this is a ball pushed into a barrel, compressing the spring, ball has initial velocity 0
here are the values,
k = 667 N/m
mass of ball = 1.5kg
ball is pushed into tube .25 meters

if u=0 is set at the mouth of the barrel, then the initial potential energy is mgh+.5kx^2?
or is it -mgh+.5kx^2? -mgh-.5kx^2?

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You get to define where the potential energy is 0. Everything above that is positive, below that negative.

what about the springs energy is that negative as well?
meaning if i set the muzzle of the gun to be potential energy 0,, is the total potential energy at the compressed initial part
-mgh-.5kx^2?

Yup, because you also defined the springs 0 potential energy mark at y=0

im trying to find the velocity once it is at y=0 (once it leaves the gun)
i am setting

$$-mgh - \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}$$

i solve this for v and i receive

$$\sqrt{\frac{2(-mgh-\frac{1}{2}kx^{2})}{m}}$$

here are the values,
k = 667 N/m
mass of ball = 1.5kg
ball is pushed into tube .25 meters

i am placing -.25 into the formula since it is below the muzzle, anyone see where i went wrong?

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U = mgh+.5kx^2

You get the negatives from the the springs compression and the height, you didn't use
-mgh-.5kx^2 THEN also set x=-.25 did you?

yes i did, i just tried again with correct negatives i think (i see where i went wrong last time)
and i got answer: 3.38m/s, wrong still though

$$mgh + \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}$$

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That would be the problem:

$$\sqrt{\frac{2(mgh+\frac{1}{2}kx^{2})}{m}}$$

Where x=-.25

Feldoh said:
Yup, because you also defined the springs 0 potential energy mark at y=0

The potential energy of the ball is +mgh. If the ball is below th etop of the barrel, h is negative, so mgh will then be negative. The elastic energy of the spring is always equal to +kx^2/2 if its unstretched position is at the top of the barrel.

jesus wrong again, i got 22.89166 m/s this time, lol

i have used
$$\sqrt{\frac{2(mgh+\frac{1}{2}kx^{2})}{m}}$$
---
$$\sqrt{\frac{2((1.5)(9.8)(-.25)+\frac{1}{2}(667)(-.25^{2}))}{1.5}}$$

o nvm i forgot the sqrt, i got correct answer now! 4.784

thank you feldoh, kamerling

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4.784 m/s once it leaves muzzle
solved now though, i didnt know how to properly set any of the initial and final energies, so i couldn't solve it initially, i got it now though.
thanks again

rsala said:
jesus wrong again, i got 22.89166 m/s this time, lol

i have used
$$\sqrt{\frac{2(mgh+\frac{1}{2}kx^{2})}{m}}$$
---
$$\sqrt{\frac{2((1.5)(9.8)(-.25)+\frac{1}{2}(667)(-.25^{2}))}{1.5}}$$

o nvm i forgot the sqrt, i got correct answer now! 4.784

Ah, that's good. I was beginning to think I just made up everything I told you since it's been a while since I've done that lol XD

## What is energy conservation?

Energy conservation refers to the practice of reducing energy consumption in order to decrease the amount of energy used and wasted. This can be achieved through various methods such as using energy-efficient appliances, reducing unnecessary energy usage, and using alternative sources of energy.

## What is the relationship between energy and mgh?

Energy and mgh (mass x gravity x height) are related through the concept of potential energy. Objects at a certain height have stored potential energy, which can be converted into other forms of energy, such as kinetic energy, as the object falls. The equation mgh represents the amount of potential energy an object has at a certain height.

## How can we conserve energy in our daily lives?

There are many ways to conserve energy in our daily lives, such as turning off lights and electronics when not in use, using public transportation or carpooling, and using energy-efficient appliances. Additionally, using renewable sources of energy, such as solar or wind power, can also help conserve energy.

## Why is energy conservation important?

Energy conservation is important for several reasons. Firstly, it helps reduce the negative impact on the environment, as the production and use of energy can contribute to air and water pollution. Secondly, it can help reduce energy costs and save money in the long run. Lastly, conserving energy helps to preserve natural resources for future generations.

## What role can individuals play in energy conservation?

Individuals can play a crucial role in energy conservation by making small changes in their daily habits, such as turning off lights and unplugging electronics when not in use, using public transportation or carpooling, and using energy-efficient appliances. Additionally, individuals can also advocate for renewable energy sources and support government policies that promote energy conservation.