University Physics 1 Work Problem

[USN2ENG]

Homework Statement

You apply a constant force F=(-68.0N)i + (36.0N)j to a 390kg car as the car travels 60.0m in a direction that is 240.0° counterclockwise from the x-axis.

How much work does the force you apply do on the car?

Homework Equations

W=Fs
W=K2-K1
W=Fcos(θ)s
F/m=a
Vf^2=Vi^2 + 2Ad

The Attempt at a Solution

Im not sure what I am doing wrong here. I have calculated the magnitude of the Force as 76.9N.

So I have tried:
76.9(cos(240))*-60 (applying the force in the same direction as the car moves) = 2307J

I also tried to use the Work energy theorem and found acceleration (76.9/390kg)=0.197m/s^2. I then used this in Vf^2=Vi^2 + 2Ad. So Vf = 4.86. I then use that in K2-K2=W=.5mVf^2-.5Vi^2. (I was assuming that Vi is 0 here, even though it didn't state from rest)

I feel like this is something simple that I am missing. Any direction would be great!

 

[Doc Al]

What's the angle between the force and the displacement?

 

[Simon Bridge]

The dot-product of the displacement with the force is the Work.
http://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation

Draw the diagram, 240 degrees anticlockwise from the +x axis is 60 degrees anticlockwise (pi/3 radiens) from the -x axis

So what is the car moves -60sin(π/3)j-60cos(π/3)i (a 2:1:√3 triangle) gives the vector:

d=(-30,-30√3)^t
F=(-68.0, 36.0)^t

You know how to evaluate a vector dot product?
http://en.wikipedia.org/wiki/Dot_product

 

[USN2ENG]

Ahhhh ok. Thanks!

I was using the angle of s and not the angle between s and F.

So the angle of F is 152.1°
240°-152.1°=87.9

so W=(76.94N)cos(87.9)(60m) = 169N

 

[Doc Al]

Much better!

 

[Simon Bridge]

Well that's the other way to do it <sniff>.

The dot product would have been:

(-30)(-69)+(-30√3)(36)=169.39J

<sulks>

 

[USN2ENG]

Haha, thanks for explaining that Simon! I guess I should have just looked at the W = F dot s to begin with and it would have made all of that a lot quicker. I am glad I have a new way of looking at that now.
Thanks Again

 

[Simon Bridge]

<perks up>
It's the way forward with computers - wait till you start working with matlab, octave, and the like. After a while it is the first thing you think of.

Basically, if you are given the angles - it's quicker to use the angles. If you are given coordinates, then the dot-product is faster and you don't have to worry about the deg/rad setting. In this case the angle of the displacement was simple, I didn't even need a calculator.