Work, potential and kinetic energy help

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Homework Help Overview

The discussion revolves around a problem involving work, potential energy, and kinetic energy related to a block sliding down an inclined plane. The block has a mass of 30 kg, and the scenario includes a frictional force of 50 N acting over a distance of 5 m with a height of 3 m.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of work done by friction and its impact on the final velocity of the block. There are attempts to apply energy conservation principles, but some participants express uncertainty about how to properly account for friction in their calculations.

Discussion Status

Participants are actively discussing their attempts and the discrepancies in their results compared to the provided solutions. Some guidance is offered regarding the interpretation of work done by friction, but there is no explicit consensus on the correct approach or final answer.

Contextual Notes

There is mention of potential errors in the provided solutions, leading to confusion among participants. The original poster expresses a lack of confidence in their understanding due to these discrepancies.

PhizKid
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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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