Showing work is path independent

1. May 27, 2013

hellocello

1. The problem:
A 5-kg block is moved from the ground to a height of 1 m via two different routes, vertically
and along an inclined plane with angle of 30° to the horizontal. How much work is done on
the block against gravity in each case?

2. Relevant equations
W=Fcosθd where θ is between the displacement and force, Fg=mg

3. The attempt at a solution
I realize the answer is that the work done is the same because it is path independent, but I don't understand why I can't show this in my calculations. Any help is greatly appreciated.

For the vertical path, I did
W=Fcosθd
W=mgcos(0)(1m)
W=(5kg)(9.8m/s^2)(1)(1m) ≈ 50 J

For the inclined path I did
W=Fcosθd where the force required to move the object along the plane is =mgsin(30°) and the angle between the displacement and force would now be 60° so
W=mgsin(30°)cos(60°)(1m) ≈12.5 J
Where I get only a fourth of the work I found above. What am I doing wrong?

(I assumed there was no friction because the question only asks about the work done against gravity)

2. May 27, 2013

TSny

For the inclined path, the displacement is not 1 m vertically. It's a certain distance along the incline.

3. May 27, 2013

hellocello

Thank you! However wouldn't that distance be 1/tan(30) ? Because then the value I get is around 20...

4. May 27, 2013

TSny

That's not quite right. Make a right triangle with one side horizontal and the other vertical. Let the hypotenuse represent the distance along the incline, and the vertical side represent the height of 1 m.

5. May 27, 2013

hellocello

Oops, it should be 1/sin(30), right? But then I'm still left with a cos(60) that doesn't cancel

6. May 27, 2013

hellocello

Oh it should be cos(0)- I forgot about that when you corrected my displacement. Thanks so much for your help!

7. May 27, 2013

TSny

You're welcome. Good work!