Welcome to the handy-dandy conservation of (mechanical) energy page. It slices and dices your HW, and simplifies difficult tasks(well, I think so, anyway). Having trouble finding velocities? You say you're sick and tired of calculating accelerations? You say to heck with Newton and his stupid F=ma? Search no more, pilgrims. You've reached the promised land.

"Conservation of Mechanical Energy":

This formula doesn't take into account friction, or so-called "dissipative" forces. A later formula will correct this. (It is not really a defect, however.) Some systems can be modeled by using this formula if the dissipative forces are small enough to be ignored. It also useful for insight into the process of energy exchange. Systems like this are sometimes referred to as closed systems: the energy they start with doesn't change.

It simply relates the change in kinetic energy to the change in potential energy (we know that Work is equal to both):

DKE = -DPE

or, expanding out the terms:

KE_i + PE_i = KE_f + PE_f

where the little "i" means initial. This equation states that the initial energy of the system is the same as the final energy. Another way to put this is that the energy of the system remains constant.

Okay, okay, I've dragged you through the derivation. Mea culpa…

So how do we actually use this info? We put in the expressions for kinetic and potential energies…

KE = (1/2)mv²

Gravitational PE = PE_grav = mgh (weight times height)

PE of a spring = PE_spring = (1/2)k(Dx)²

Using these terms we can now write (using, for example, gravitational potential energy),

KE_i + PE_i = KE_f + PE_f

(1/2)mv_i² + mgh_i = (1/2)mv² + mgh

And now we can solve real problems dealing with altitudes and velocities.

The key is to recognize when v=0 and/or when the potential energy, mgh, can be set to zero (a level which you are free to choose. How? You can choose the ground to be zero, or a ceiling, or a desk top. In some of these cases the PE will be negative. The formula is consistent as long as you use the same reference level and coordinate system)

Let's try one example:

A roller coaster is moving at 5 m/s and is 30 m above the ground. What is the velocity when it reaches the ground?

v_i = 5 m/s

h_i = 30m

h_final = 0

so…

(1/2)mv_i² + mgh_i = (1/2)mv² + mgh

notice that mass conveniently cancels out. (Galileo was right!)

(1/2)v_i² + gh_i = (1/2)v² + 0

v² = v_i² + 2gh

v² = (5 m/s)² + 2(9.8 m/s²)(30m)

v = 24.8 m/s