• Support PF! Buy your school textbooks, materials and every day products Here!

University Physics 1 Work Problem

  • Thread starter USN2ENG
  • Start date
  • #1
110
0

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!
 

Answers and Replies

  • #2
Doc Al
Mentor
44,892
1,143
What's the angle between the force and the displacement?
 
  • #3
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
Last edited:
  • #4
110
0
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
 
  • #5
Doc Al
Mentor
44,892
1,143
Much better!
 
  • #6
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
Well that's the other way to do it <sniff>.

The dot product would have been:

(-30)(-69)+(-30√3)(36)=169.39J

<sulks>
 
  • #7
110
0
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
 
  • #8
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
<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.
 
Top