Homework Help: Work done by a constant force

1. Jul 1, 2005

nissanfreak

Work done by a constant force!!!

A small plane tows a glider at constant speed and altitude. If the plane does 2.00 X 10^5 J of work to tow the glider 145m and the tension in the tow rope is 2560N, what is the angle between the tow rope and the horizontal?

This is one of my homework problems. I think that I might be using the wrong formula for this. What formula should I use? And If its possible could someone please show me how to solve this step by step I really want to understand this stuff, so any and all help would be greatly appreciated!!!

2. Jul 1, 2005

Staff: Mentor

Why don't you show us what you've done so far?

Hint: The glider moves horizontally. So what component of the rope tension is doing the work on the glider?

3. Jul 1, 2005

nissanfreak

I know that the answer is 57.4 degrees, but I just cant seem to get it? The formula I am using is W=Fd(cos phada) And I am trying to solve for phada. Can someone please work out this problem step by step for me? I want to learn!!!

4. Jul 1, 2005

Astronuc

Staff Emeritus
or Force = Work / distance.

Resolve the force applied horizontally and vertically. Which applies to the work being done?

5. Jul 2, 2005

Knavish

Since the plane is pulling the glider horizontally (and not vertically), the horizontal force corresponds to the work being done. Thus F=2560cos(theta).

I think you can solve from there.

6. Jul 2, 2005

Staff: Mentor

Realize that you are given values for W, F (that's the tension in the rope), and d. All you need do is solve for cos(theta). Then use your calculator to find theta.

To solve for cos(theta): Divide both sides of your formula by Fd.

7. Jul 2, 2005

nissanfreak

Well ive tried solving it like that but the answer I keep getting is 5.38793103e-11. Then to get rid of the e-6 I move the decimal over to the right 11 spaces and I then get 54 degrees as my answer. But the books answer is 57.4 degrees. What am I doing wrong?

This is how I had the equation looking .0000200/(2560)(145)=cos theta

8. Jul 2, 2005

Staff: Mentor

The work is given as 2.00 X 10^5 J = 200,000 J. (Not 2.00 X 10^-5 J = 0.0000200 J.) You messed up the exponent.