# Question on energy

## Homework Statement

A .444kg block, slides down a .888m ramp, at an angle of 33.3°. The block starts from rest, and reaches a speed of 2.22 m/s. How much energy is lost?

## Homework Equations

If I'm thinking correctly, I'd use the work equation...
W=ΔK+ΔUg+ΔUs
ΔK=(1/2)m*vf^(2) - (1/2)m*vi^(2) (The second part of this drops out, because the initial velocity is 0)
Oh, and there's no spring involved, so ΔUs drops out.

## The Attempt at a Solution

So, the stated equation can be re-written as...
(1/2)m*vf^2 +m*g*Δy (ΔY=Δxsin∅)
Then...
(1/2)(.444kg)(2.22^(2)m/s^(2)) + (.444kg)(9.80m/s^2)(.888m sin(33.3°))

The answer I get, assuming I'm right is 3.22 J? Am I getting anywhere with this, or am I completely off?

## Answers and Replies

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Hi Timebomb,

can you clear up something in the question for me? When the question asks "how much energy is lost", what does that mean?

Hi Timebomb,

can you clear up something in the question for me? When the question asks "how much energy is lost", what does that mean?
I don't know. That's why I'm asking. That's exactly what the question says, on my assignment.

I don't know. That's why I'm asking. That's exactly what the question says, on my assignment.
So if your answer is around 3 Joules, where is that energy going? Through what process is the energy lost?

Oh I see what the question is saying. Okay, I also see your mistake. your potential energy is the maximum energy that you had when the block was at rest. So you shouldn't be adding the kinetic energy to it, you should be subtracting the kinetic energy from it.

Basically the energy lost is equal to all of the potential energy that didn't get converted to kinetic energy.

So if your answer is around 3 Joules, where is that energy going? Through what process is the energy lost?
Maybe the 3.2 Joules is the energy lost/given off, by the block sliding down the .888m ramp? If that isn't right, then I don't know what is.

Maybe the 3.2 Joules is the energy lost/given off, by the block sliding down the .888m ramp? If that isn't right, then I don't know what is.
Yeah, you're right, it must be friction. did you read my other post?

Yeah, you're right, it must be friction. did you read my other post?
Yea, it's the friction. Because the next part of the problems says, "If the energy is lost to frictional heating, what is the magnitude and direction of the frictional force." I'm just curious if the number I got was right.