Find Work Done on Homework Problem Involving Constant Force & Vector of PQ

In summary, the conversation discusses finding the work done when a constant force is applied and the point of application moves along a line segment from point P to point Q. The hint given is to find a vector b that represents PQ. The conversation also mentions using components and the dot product to solve the problem. The final answer is determined to be -36, as the force is in the opposite direction of the displacement. The conversation concludes with the original poster thanking the person who provided the explanation and apologizing for any incorrect terminology.
  • #1
TrueStar
95
0

Homework Statement


If c represents a constant force, find the work done if the point of application of c moves along the line segment from P to Q.
(Hint: Find a vector b = (b1, b2) such that b is a vector of PQ)

c = -i+7j P(-2,5) Q(6,1)

Homework Equations


Components...but I am not sure. Dot Product.


The Attempt at a Solution


We went over components and dot product in class, but not how it relates to work. I'm not even sure to begin with this one.
 
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  • #2
TrueStar said:
We went over components and dot product in class, but not how it relates to work.

Hi TrueStar! :smile:

work done = force "dot" displacement (of the point of application of the force). :wink:
 
  • #3
Thanks. :) I've researched it myself..I tried the problem and the answer I got was 32. I found many examples where one of the vectors was at the origin, but not two vectors with different components.

This is what's throwing me off. Then again, maybe I got the right answer. I don't know yet.
 
  • #4
Maybe you've used the wrong vector for PQ.

Show us how you got 32. :smile:
 
  • #5
Thanks Tim:

I made a mistake. I think it's 36. I kind of did it my own way. I knew how to get the answer if one of the vectors was at (0,0). So, I picked -2,5 and decided that I needed to add two to x and subtract 5 from y. I did this to the other vector which changed its components from 6,1 to (8,-4).

I then used dot product with (8,-4) and -i+7j and got -36. I noticed similar problems to this had positive answered, so I put down 36.

I know..I'm horrible. I'd rather put something down instead of nothing at all.
 
  • #6
TrueStar said:
Thanks Tim:

I made a mistake. I think it's 36. I kind of did it my own way. I knew how to get the answer if one of the vectors was at (0,0). So, I picked -2,5 and decided that I needed to add two to x and subtract 5 from y. I did this to the other vector which changed its components from 6,1 to (8,-4).

I then used dot product with (8,-4) and -i+7j and got -36. I noticed similar problems to this had positive answered, so I put down 36.

I know..I'm horrible. I'd rather put something down instead of nothing at all.

D'oh! :rolleyes:

You got it right, and then you changed it so as to "conform".

It is -36 … this particular one comes out negative because the force is in (roughly) the opposite direction to the displacement …

eg what work does gravity do when a roller-coaster goes up? Obviously, negative! :smile:

This is the way to do it …

(btw, don't talk about "changing the components" of a vector … you can't do that to vectors (well, unless you're changing the frame of reference, or the basis)! … talk about subtracting them … that is the official word … they form a "vector space" under addition subtraction and scalar multiplication :wink:)

P = (-2,5) Q = (6,1), so PQ = Q - P = (8,-4) or 8i - 4j

c = (-1,7) or -i + 7j

Then you can find c.PQ either by doing (8,-4).(-1,7) or (8i - 4j).(-i + 7j),

(both are perfectly valid)

and either way it's -8 -28 = -36 :wink:
 
  • #7
Oh wow..I just can't believe I got as much correct as I did! The instructor never talked about the concept of work in relation the the topic (components). Our textbook has examples of this where one was at the origin, but not like this one.

I apologize for my incorrect terminology, this is all new to me and I have yet to take a physics class.

Thank you so much for helping me out and explaining everything so clearly. I really appreciate it!
 

1. What is work done on a homework problem involving constant force and vector of PQ?

The work done on a homework problem involving constant force and vector of PQ is the product of the magnitude of the force and the displacement of the object in the direction of the force. In other words, it is the amount of energy transferred to the object to move it from its initial position to its final position.

2. How do I calculate the work done on a homework problem involving constant force and vector of PQ?

To calculate the work done on a homework problem involving constant force and vector of PQ, you can use the formula W = F * d * cosθ, where W is the work done, F is the magnitude of the force, d is the displacement, and θ is the angle between the force and the displacement.

3. What is the unit of measurement for work done on a homework problem involving constant force and vector of PQ?

The unit of measurement for work done is joules (J) in the International System of Units (SI). In some cases, it may also be measured in newton-meters (N⋅m) or foot-pounds (ft⋅lbf).

4. Can the work done on a homework problem involving constant force and vector of PQ be negative?

Yes, the work done on a homework problem involving constant force and vector of PQ can be negative. This occurs when the force and the displacement are in opposite directions, resulting in the displacement being negative. This can also be interpreted as the object losing energy instead of gaining it.

5. How does the angle between the force and the displacement affect the work done on a homework problem involving constant force and vector of PQ?

The angle between the force and the displacement affects the work done on a homework problem involving constant force and vector of PQ through the cosine function in the formula W = F * d * cosθ. When the angle is 90 degrees, the cosine is equal to 0, resulting in no work being done. When the angle is 0 degrees, the cosine is equal to 1, resulting in maximum work being done.

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