Kinetic Energy Formula Expansion / Wnet relation.

1. Nov 19, 2007

Senjai

[SOLVED] Kinetic Energy Formula Expansion / Wnet relation.

1. The problem statement, all variables and given/known data
I was given two formulas for Kinetic Energy from my physics teacher, can anyone explain were they derive from? I like to know were they come from, not just how to use them

2. Relevant equations

$$E_k = \frac{1}{2}mv^2$$

$$W_{net} = \Delta E_k$$

$$\Delta E_g = mgh$$

3. The attempt at a solution

none, i just would like to know were they come from, i know they have to come from some formula i already know. How does Work done by a net force differ from work done by an arbitrary force?
$$W_{net} = ( \frac{1}{2}m_1v_2^2) - ( \frac{1}{2}m_1v_1^2)$$ << is that right and how does work equal energy?

Last edited: Nov 19, 2007
2. Nov 19, 2007

Mindscrape

Kinetic energy is simply defined that way. Similarly, potential energy is defined as the work done to bring a particle to a certain point. Potential energy comes from work. The work to bring a particle of mass m to a point above the earth is

$$W = \int \mathbf{F} \cdot d\mathbf{r}$$

since gravity can be approximated, in a flat earth approximation, to only act in the up and down direction, the only work done can be in that same direction, i.e.

$$W = \int F dz[/itex] where I have defined the z direction to be up and down direction. The gravitational force close to the earth is a constant, mg, so the work to bring a particle to a point above the earth is [tex] W = mg \int_{Height1}^{Height2} dz = mg(h_2 - h_1) = mgh_2 - mgh_1 = \Delta E_g$$

Most of the time you define your first height to have a zero potential energy, so that you just have $E_g = mgh_2 = mgh$.

From energy conservation you know that the change in the potential energy must equal the change in kinetic energy, so..

$$W_{net} = \Delta E_g = \Delta E_k$$

Different forces have different potential energies, but if you know the difference in kinetic energies you will know the work done (given a conservative system), so you would be right.

3. Nov 19, 2007

Senjai

I was pondering something, if you picked up a rock, carried it up to the top of the hill, youd transfer a certain amount of Kinetic energy to the rock as your doing work against gravity, which is a conservative system. from my understanding work must be given back in some way in a conservative system, so you bring the rock to the top of the hill, and somehow it rolls down to the origin you picked it up at, so the Kinetic Energy would be 0 right? same with displacent E_k1 = E_k2 so when E_k2 - E_K1 would equal 0 change in kinetic energy right? meaning that no work was exerted... how is this possible, you had to exert work to get it up there, and you didnt get the work back.. this is how im lost.

4. Nov 19, 2007

Senjai

Same with if you pick up something, and drop it, what if it goes past its original origin e.g pick it up off the floor, drop it, and it goes through the floor. in this case energy wouldn't be conserved right?

EDIT: on another note, so..

$$W_{net} \equiv \frac{1}{2}mv^2$$

equals by definition?

Last edited: Nov 19, 2007
5. Nov 19, 2007

Mindscrape

Hmm, not quite sure what you are saying, but here is what happens.

You carry a rock to the top of a hill, you have done work to get that rock up there. The rock now has potential energy equal to mgh (if we define the zero potential to be at the initial height). The kinetic energy when you first drop it at height h is zero, and the potential energy when it reaches the bottom of the hill is defined as zero. So from $E_i = E_f$ we have $mgh = \frac{1}{2}mv^2$. The work you have done to get the rock to the top now gives the rock kinetic energy when it reaches its initial state.

6. Nov 19, 2007

Senjai

I see, thanks much!

7. Nov 19, 2007

Mindscrape

You can only say

$$W_{net} \equiv \Delta U$$

where U is the potential energy

If the system is conservative you can further say that

$$\Delta U = \Delta T$$

where T is the kinetic energy.

So, you could claim (in a conservative system)

$$W_{net} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$$

Example of dropping something that goes below floor
$$E_i = E_f$$

$$mgh = -mgh_2 + \frac{1}{2}mv^2$$

the particle went below the zero potential, so the final energy will have a nonzero potential energy (in fact it is below the zero potential, so it is negative).

8. Nov 19, 2007

Senjai

no clue how much that helps, thank you so much..

ill be back with more kinetic questions soon haha..