Calculating Work Done by a Steady Wind in a Given Direction

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Sneakatone
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a strong steady wind provides a force of 170 N in a direction 30 degrees east of north on a person. If the person walks first 110 m north and then 190 m east what is the total work done by the wind?

I used W=FDcosθ
170(110)cos(30)=16194.67J
170(190)cos(30)=27972.62

should 110 m north be 90 degrees?
 
Sneakatone said:
should 110 m north be 90 degrees?

Yes, you can take north to mean 90 degrees, but that is 90 degrees from the positive x-axis. So, if you do that, then you have to take all of your angles from the positive x-axis.

What you want for your θ is the angle between the Force and the displacement. You have two displacement vectors.

What is the angle between the first "walk" and the force of the wind. And what is the angle between the second "walk" and the wind.
 
would this picture be right?
 
Attachments
  • Hand-drawn sketch of a triangle with a horizontal line labeled 140 at the top, a vertical line on the left labeled 110, and a small angle marked at the bottom vertex between the vertical and slanted sides, sitting on a horizontal ground line.
    py.png
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I would do 2 free body diagrams. Think of the this as two problems. "How much work is done by the wind during your 110m walk?" and "How much work is done by the wind on your 190m walk?"

You have the right idea in the orginal post. You just need to find the correct angles.
So, what are your angles?
 
sorry for the bad paint writing
 
Attachments
  • Hand-drawn coordinate axes with a right triangle in the first quadrant: a line from the origin at an angle to a vertical side, with a small angle arc at the origin and a handwritten label near the angle.
    p2.png
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  • Handwritten sketch of a coordinate axis cross with a right triangle in the first quadrant, the angle at the origin labelled 30 degrees and the vertical side labelled 150.
    p1.png
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I attached a FBD for the first part. See what you can take from it and apply it to a FBD for the 2nd part. and Then find the work done in each instance.

Notice your problem says30 degrees due EAST of NORTH.
 
Sorry, here it is.
 
Attachments
  • Handwritten green diagram on a whiteboard for problem 1: north and east axes meet at a point, with displacement d = 3.10 m drawn along north and force F = 170 N drawn at 30 degrees from north.
    IMAG0550.jpg
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the angle for part 2 is 60
 
Attachments
  • Handwritten sketch of a pair of perpendicular axes with a ray drawn at 60° above the horizontal, labelled 60° and 190 m along the horizontal axis.
    p3.png
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That looks good, assuming I'm looking at it right. I understand you're using paint do draw these so it's a little tough. Just be sure you're labeling everything correctly (see my diagram) when you draw them on your paper.

So what you do get for your work?
 
170(110)cos(30)=16194.67J
170(190)cos(60)=16150 J
I added em and it was correct.
Thank you soo much again! I guessed I didnt not really know what east of north along with it surrounding angles ment until now.
 
Yeah, wording can be tricky sometimes, just have to pay real close attention.