Work, potential and kinetic energy help

AI Thread Summary
The discussion revolves around a physics homework problem involving a 30 kg block sliding down an inclined plane, where participants attempt to calculate the work done by friction and the final velocity. Initial calculations for work done by friction yielded -250 J, which was deemed incorrect, with the correct answer being -150 J. Participants struggled to incorporate friction into their potential energy calculations, leading to discrepancies in their final velocity results. Despite multiple attempts, the calculated velocities did not match the expected solution of 7.3 m/s, raising concerns about the accuracy of the provided answers. The conversation highlights the challenges of understanding concepts when faced with potential errors in textbook solutions.
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Homework Statement


A 30 kg block is slid down an inclined plane 5 m long with a height of 3 m from the floor. If the force of friction is 50 N, find the work done by the friction and the final velocity at the end of the plane.


Homework Equations


W = F*d
mgh = (1/2)mv^2

The Attempt at a Solution


50 N * 5 m * cos(180) = -250 J

Incorrect

30 kg * 9.8 m/s^2 * 3 m = (1/2) * 30 kg * v^2
882 = 15*v^2
58.8 = v^2
v = 7.6681 m/s

Incorrect

I don't know how else to approach this.
 
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PhizKid said:
50 N * 5 m * cos(180) = -250 J

Incorrect
Nothing wrong with this. They probably wanted just the magnitude of that work.

30 kg * 9.8 m/s^2 * 3 m = (1/2) * 30 kg * v^2
882 = 15*v^2
58.8 = v^2
v = 7.6681 m/s
You forgot about the friction.
 
The answer says -150 J so I guess that's an error then.

How do I account for fricton in potential energy?
 
PhizKid said:
The answer says -150 J so I guess that's an error then.
Looks like they used the 3 m distance by mistake.
How do I account for fricton in potential energy?
Initial mechanical energy + work done by friction = Final mechanical energy
 
So then I get (30 kg * 9.8 m/s^2 * 3 m) + (-250 J) = (1/2) * 30 kg * v^2

I get 6.49 m/s which is also incorrect (solution is 7.3 m/s).
 
PhizKid said:
So then I get (30 kg * 9.8 m/s^2 * 3 m) + (-250 J) = (1/2) * 30 kg * v^2

I get 6.49 m/s which is also incorrect (solution is 7.3 m/s).
I agree with your answer. Even using their incorrect value for the work done by friction you won't get their answer.

What book is this from?
 
Something our professor wrote :-p

It's so intimidating because I feel like I'm not getting the concepts when it just turns out the solutions are incorrect...so I'm still not confident enough to say that my answer is correct and the answer key is not
 
It is unfortunate when the professor makes errors, but it happens.

Your textbook should have problems for you to work on. Often some of the answers are given.

You could also supplement your text with a problem book, such as a Schaum's Outline.
 
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