Work, potential and kinetic energy help

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.

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.

Doc Al
Mentor
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

The answer says -150 J so I guess that's an error then.

How do I account for fricton in potential energy?

Doc Al
Mentor
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).

Doc Al
Mentor
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 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

Doc Al
Mentor
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.