Calculating Work Done by a Steady Wind in a Given Direction

In summary, the strong steady wind provided 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, the total work done by the wind is 16194.67J.
  • #1
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?
 
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  • #2
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.
 
  • #3
would this picture be right?
 

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  • #4
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?
 
  • #5
sorry for the bad paint writing
 

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  • #6
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.
 
  • #7
Sorry, here it is.
 

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  • #8
the angle for part 2 is 60
 

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  • #9
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?
 
  • #10
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.
 
  • #11
Yeah, wording can be tricky sometimes, just have to pay real close attention.
 

1. What is the formula for calculating total work done?

The formula for calculating total work done is force x distance, where force is measured in newtons (N) and distance is measured in meters (m).

2. How do you find the force in a work equation?

The force in a work equation can be found by multiplying the mass of an object by its acceleration, which is measured in meters per second squared (m/s²).

3. What is the unit of measurement for work?

The unit of measurement for work is the joule (J). One joule is equal to one newton-meter (Nm) or one kilogram-meter squared per second squared (kg·m²/s²).

4. Can work be negative?

Yes, work can be negative if the force and displacement are in opposite directions. This indicates that the object is losing energy or moving in the opposite direction of the applied force.

5. How do you find the total work done when multiple forces are involved?

To find the total work done when multiple forces are involved, you can use the principle of superposition and add together the work done by each individual force. Alternatively, you can use the work-energy theorem, which states that the total work done on an object is equal to its change in kinetic energy.

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