Not sure how to plug in numbers for Work Energy Theorem

[CiCi]

1. The Problem Stament, all variables and given data

a 15 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20 degrees with the horizontal. if there is a constant force of kinetic friction of 25 N between the crate and ramp, what kinetic energy would the crate have at the bottom of the ramp?

2. Needed equations

Work = PE + KE + Energy Lost
PE= mgh
Height = Sinθ * Length
KE= (1/2) m*v^2

3. Attempt at solution

okay so I know how to start out with finding PE

PE= mass* gravity* height = 15 * 9.8 * height

Find Height
Height= Sin (20°) * 2.0= .684

PE= 15 * 9.8 * .684 = 100.5 Joules

What's throwing me off is the kinetic friction of 25 N, I'm not sure what to do with it and where to plug it into to

 

[Orodruin]

How can you include the work done by the friction into your energy considerations?

 

[CiCi]

How can you include the work done by the friction into your energy considerations?

Is there another equation I need? Because I'm not sure

 

[CWatters]

What equation relates force and distance to work?

 

[CiCi]

What equation relates force and distance to work?

Force = distance / time ?

 

[CWatters]

I said force, distance and work.
(Not time)

 

[CiCi]

I said force, distance and work.
(Not time)

Work = force * distance ?

 

[CWatters]

That's the one.

 

[CiCi]

That's the one.

So
Work = 25 N * 2.0 m
= 50 Joules
What does this do?

 

[betamu]

That describes the amount of energy lost to friction. You know that this is the only way energy is lost during the motion of the object. Therefore you can say that the final energy of the box will be the initial energy minus the energy lost to friction.

 

[CWatters]

+1

You have a system where energy changes from PE to KE and heat. Apply the relevant equation you posted that invokes conservation of energy.

 

[kuruman]

Work = force * distance ?

It's more than that.
For a constant force, the work is given by $W=Fd\cos \theta$, where $F$ = magnitude of the force, $d$ = magnitude of the displacement and $\theta$ = angle between force vector and displacement vector. Work is "force times distance" only if the force and the displacement are in the same direction. This is not the case here.

 

[CWatters]

It's more than that.
For a constant force, the work is given by $W=Fd\cos \theta$, where $F$ = magnitude of the force, $d$ = magnitude of the displacement and $\theta$ = angle between force vector and displacement vector. Work is "force times distance" only if the force and the displacement are in the same direction. This is not the case here.

That's correct but I hope it doesn't confuse cici. In this case the force and displacement arent in the same direction but are parallel so trig isn't really needed to solve.

 

[kuruman]

That's correct but I hope it doesn't confuse cici. In this case the force and displacement arent in the same direction but are parallel so trig isn't really needed to solve.

Well, the title of the thread mentions use of the work-energy theorem which implies calculation of the sum of all the works. None of the acting forces (gravity, friction and normal force) are in the direction of the displacement. I hope @CiCi does not get confused either, but he has to understand how the more general definition of work is implemented.