Calculating Work Done by a Constant Force

[opticaltempest]

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

 

[tony873004]

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.

 

[opticaltempest]

I made a few changes to my sketch. How does that look?

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


Thanks

 

[tony873004]

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.