Calculating Work Done by a Constant Force

Click For Summary

Homework Help Overview

The discussion revolves around calculating the work done by a constant force, specifically in the context of a force diagram and its components. Participants are examining the correctness of a solution related to a physics problem involving force and distance.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster seeks validation of their solution and presents a force diagram. Some participants question the accuracy of the diagram and the representation of force components. There are discussions about the relationship between force, distance, and the calculation of work.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's sketches and calculations. Some guidance has been offered regarding the correct representation of force components and the need to clarify the force diagram. Multiple interpretations of the problem setup are being explored.

Contextual Notes

Participants note potential issues with mixing units in the diagrams, as well as the necessity of accurately representing force components in the context of the problem.

opticaltempest
Messages
135
Reaction score
0
I have the following problem,

http://img236.imageshack.us/img236/1684/problem7ke.jpg

http://img236.imageshack.us/img236/6213/image8oa.jpg

Is my solution correct?

http://img235.imageshack.us/img235/9374/solution1ez.jpg


Thanks
 
Last edited by a moderator:
Physics news on Phys.org
Your drawing is not right, but your answer is.

AC is not 50 feet, it the x-component of the 25 pounds of force. You computed it correctly with cos(20)*25, and then multiplied it by distance to get work.
 
I made a few changes to my sketch. How does that look?

http://img92.imageshack.us/img92/4974/100018at.jpg


Thanks
 
Last edited by a moderator:
You get the concept, but you're not going to have feet and pounds in the same diagram. You can't add them together. Just get rid of point E. AC is all you need to complete your force diagram. W=F*D. You computed force correctly. You multiplied it by distance and got work. Now just get distance out of your force diagram. It doesn't belong there. If you want, you can draw a separate 1-dimensional diagram showing distance on the x-axis, but that is not necessary. Just get rid of your CE vector and you're finished.
 

Similar threads

Force on electron from current-carrying wire
  • · Replies 2 ·
Replies
2
Views
4K
Prof said one thing, book says another. R/L source-free circuit, pic included
  • · Replies 2 ·
Replies
2
Views
2K
Work done by a balloon that is rising
  • · Replies 8 ·
Replies
8
Views
1K
About Work done when velocity is constant
  • · Replies 12 ·
Replies
12
Views
4K
Work Pulley Problem with constant speed
  • · Replies 2 ·
Replies
2
Views
1K
Work done to insert a point charge 𝑞 at the center of a conducting sphere
  • · Replies 12 ·
Replies
12
Views
2K
How does the calculation of work change for a rotating or non-rigid object?
  • · Replies 4 ·
Replies
4
Views
3K
Work done by a force down a ramp
  • · Replies 11 ·
Replies
11
Views
2K
How can "work" done by a force be negative?
  • · Replies 19 ·
Replies
19
Views
2K
Work done on a pendulum by gravity
  • · Replies 7 ·
Replies
7
Views
3K