Energy Loss in .444kg Block Sliding Down .888m Ramp

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A .444kg block slides down a .888m ramp at a 33.3° angle, starting from rest and reaching a speed of 2.22 m/s. The energy lost during this process is calculated using the work-energy principle, which involves potential and kinetic energy. The initial potential energy is not fully converted to kinetic energy, indicating energy loss primarily due to friction. The calculated energy loss is approximately 3.22 Joules, which is attributed to frictional heating as the block moves down the ramp. Understanding this energy loss is crucial for solving related problems, such as determining the frictional force involved.
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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?
 
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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?
 
dacruick said:
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.
 
Timebomb3750 said:
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.
 
dacruick said:
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.
 
Timebomb3750 said:
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?
 
dacruick said:
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.
 
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