What is it called when one type of energy is changed into another?

Answer 1

Energy has many different forms, and the energy in a system changes forms according to the laws of physics. Here is an example of energy changing form from "potential" to "kinetic."

Consider dropping a ball that was held in the air some distance "h_top" above a surface. Let's name some variables:

"h" (in meters), is the height of the ball after you drop it at any given time. So if you drop the ball and wait a little while, "h" will no longer be equal to h_top,but will have decreased a small amount. Got it?

"g" is Earth's local gravitational field, in m/s/s (this is always 9.81 m/s/s), "v" is the ball's speed at any given time and "m" is the mass of the ball in kilograms. The product of m*g*h is called the ball's "gravitational potential energy" ("PE"). The product of half of "m" and the speed of the object (in m/s) squared is called the ball's "kinetic energy" ("KE").

PE = mgh

KE = (1/2)m*(v^2)

If we drop the ball and take a snapshot of it at ANY point on the way down, we can (in an ideal world) evaluate these two energies, add them, and get a number called the ball's "mechanical energy."

ME = PE + KE

The idea of energy conservation here is that mechanical energy comes out to the same value no matter when we take our snapshot of the ball during its descent. So if ME comes out to be 5 or something at the top, it will be 5 at the bottom and 5 at every point in between!

So, if we're clever, we can choose a point along the ball's path that makes the math easy to evaluate ME: let's choose the point when the ball hasn't been dropped yet. This is convenient to plug in, because presumably you know its initial height "h_top", and its velocity is 0 (as you haven't dropped it yet). Since velocity is 0, the whole KE term = 0. You can skip the plugging in, squaring, and multiplying. Easy.

So let's evaluate ME at the point when the ball has not been dropped yet (which we'll call "ME_top"). Note that "g" and "m" do not depend on where the ball is, but "h" (the ball's current height above the surface) does, and "v" (the ball's speed) does, so we need to plug "h_top" and "v_top" into this equation to find ME_top.

ME_top = PE_top + KE_top

ME_top = mg(h_top) + (1/2)m*(v_top^2)

v_top = 0 (because we haven't dropped it yet, so it isn't moving)

ME_top = mgh_top + 0

ME_top = mgh_top

So ME_top is equal to whatever m*g*h_top comes out to. But remember, "ME" for the ball is the same no matter what point in time you choose along the ball's drop, so ME_top is equal ME_bottom or ME_middle or ME_anywhere.

Let's pretend we don't know this and try to evaluate ME again at the bottom:

ME_bottom = mg(h_bottom) + (1/2)m*(v_bottom^2)

h_bottom = 0 (height above the surface is 0)

ME_bottom = 0 + (1/2)m*(v_bottom^2)

ME_bottom = (1/2)m*(v_bottom^2)

Here, we have to stop, because we don't necessarily know "v_bottom" (the speed of the ball at the bottom of its descent). But we know that if ME is always the same, then the value we got for it at the top should be equal to the value we got for it at the bottom. That means:

ME_bottom = ME_top

(1/2)m*(v_bottom^2) = mg(h_top)

This last equation allows us to solve for v_bottom. This is a common physics problem. Also, this result ultimately answers your question!

The equation clearly says that the potential energy found at the top exactly equals the kinetic energy found at the bottom.

At the top, all of the energy was potential, and at the bottom, all of it was kinetic. In effect, by moving through Earth's gravitational field, energy changed forms - it went from potential to kinetic!

Answer 2

That is kinetic energy, being by which energy is transform into another form. Like the suns rays, transform that into electrical energy.

Answer 3

This process is called energy conversion, where energy changes from one form to another. This occurs in various systems and processes, such as when potential energy is converted to kinetic energy.

Answer 4

Energy conversion