# University Physics 1 Work Problem

## 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?

W=Fs
W=K2-K1
W=Fcos(θ)s
F/m=a

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

Simon Bridge
Homework Helper
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

Last edited:
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
Mentor
Much better!

Simon Bridge
Homework Helper
Well that's the other way to do it <sniff>.

The dot product would have been:

(-30)(-69)+(-30√3)(36)=169.39J

<sulks>

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. Im glad I have a new way of looking at that now.
Thanks Again

Simon Bridge