Showing work is path independent

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hellocello
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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?

Homework Equations


W=Fcosθd where θ is between the displacement and force, Fg=mg


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)
 
on Phys.org
hellocello said:
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

For the inclined path, the displacement is not 1 m vertically. It's a certain distance along the incline.
 
Thank you! However wouldn't that distance be 1/tan(30) ? Because then the value I get is around 20...
 
hellocello said:
However wouldn't that distance be 1/tan(30) ?

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.
 
Oops, it should be 1/sin(30), right? But then I'm still left with a cos(60) that doesn't cancel
 
Oh it should be cos(0)- I forgot about that when you corrected my displacement. Thanks so much for your help!