Calculating Work and Mechanical Energy - Block on Ramp

1. The problem statement, all variables and given/known data

Starting from rest, a 5.0 kg block slides 2.5 m down a rough 30.0 degree incline in 2.0 seconds. Determine the following:

a. work down by the force of gravity
b. the mechanical energy lost due to friction
c. the work done by the normal force between the block and the incline

2. Relevant equations
I think Wnet = Fnetd(cos thata) is relevant.
KE = 1/2mv^2
and PEg = mgh

3. The attempt at a solution

I started by drawing a force diagram...49.1 N pointing downards and a 30 degree angle drawn in. I know that I need to figure out Forcenet but I'm not sure how. I would attempt more of a solution but I'm honestly just that bad at physics. I've worked all week and I have an exam on Friday, but I'm still clueless. Help is really needed here. I want to understand.
Hi mimib1230. Welcome to PF.

Try starting by spliting the weight of the mass into components. From this you'll have a force down the slope (but which one?) and this will then allow you to substitute the value into an equation (but which one?). See what you make of that. I will not give you anymore as I read that you want to understand. Small hints at a time. :biggrin:

The Bob
Fparallel = Fgsin30 = (49.1 N ) sin 30 = 24.6 N

Ffriction? I don't have the coefficient of friction, so how can I figure out my Fnet?
Because you know the force needed to overcome friction down the slope and also the normal reaction so could work out the coefficient of friction. However, you don't really need this to work it out. As you know the other information in your question you can find the work done by comparing initial and final answers to this problem.

The Bob
A lot of the time, I don't know exactly where to start (at all). My instinct here was to draw a force diagram, figure out friction, do a billion different things to get Forcenet.

But actually I asked my teacher and all I needed was to draw a triangle to figure out the height for part A.

So h= 1.25 from the ground because 2.5sin30 = 1.24 m.

PEgravity = (5.0 kg)(-9.81 m/s/s) (-1.25 m) = 61.31 J

delta x = 1/2 (vi + vf) time
-2.50 m = 1/2 (vf) (2.00 s)
vf = 2.5 m/s

KE = 1/2 (5.0 kg) (-2.5 m/s)2 = 15.625 J

PEgravity - KE = energy lost due to friction
61.31 J - 15.625 J = 47.685 J

Is part C simply "no energy?" It is only moving up and down via the path of the incline, not through the air, right?
hello, i have a question kinda like that but different.

it says a football player pushes a box that wieghs 300n up a ramp that is 6 meters long with a force of 150n while the person on the box is 3 meters high. what is the actual work done?

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