How can you use the line integral to find the work done by a conservative force?

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SUMMARY

The discussion focuses on calculating the work done by a man carrying a 25-lb can of paint up a helical staircase surrounding a 90-ft high silo with a radius of 20 ft. The work is determined using the line integral of the dot product of the force vector and the path's derivative. The correct approach to find the work done against gravity is to apply the formula W = mgh, where m is the total weight (185 lbs) and h is the height (90 ft), yielding a total work of 16,650 ft-lbs.

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  • Knowledge of conservative forces and gravitational force
  • Familiarity with parametric equations for helical paths
  • Ability to compute dot products of vectors
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fk378
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Homework Statement


A 160-lb man carries a 25-lb can of paint up a helical staircase that encircles a silo witha radius of 20 ft. If the silo is 90 ft high and the man makes exactly three complete revolutions, how much work is done by the man against gravity in climbing to the top?


Homework Equations


W= line integral of dot product of F(r(t)) -dot- r'(t)



The Attempt at a Solution


My r(t)= (20cost, 20sint, 90)
r'(t)=(-20sint, 20cost, 0)
t is between 0 and 6pi
I know that gravity is -9.8 in the k direction. However I don't know what to use for my vector field...or if I should be using a different r(t). At first I tried using <0,0,-9.8> -dot- <-20sint,20cost,0> but obviously that just gives me zero.
 
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The path taken is that described of a helix. How would you parametrise a helix? It should have non-zero components for all the i,j,k directions.
 
fk378 said:

Homework Statement


A 160-lb man carries a 25-lb can of paint up a helical staircase that encircles a silo witha radius of 20 ft. If the silo is 90 ft high and the man makes exactly three complete revolutions, how much work is done by the man against gravity in climbing to the top?


Homework Equations


W= line integral of dot product of F(r(t)) -dot- r'(t)



The Attempt at a Solution


My r(t)= (20cost, 20sint, 90)
r'(t)=(-20sint, 20cost, 0)
t is between 0 and 6pi
I know that gravity is -9.8 in the k direction. However I don't know what to use for my vector field...or if I should be using a different r(t). At first I tried using <0,0,-9.8> -dot- <-20sint,20cost,0> but obviously that just gives me zero.

It's a conservative force so the answer is simply mgh.
You may take the total displacement vector dotted with minus the force of gravity and you get of course mgh.

Do you have to prove it with an integral? I don't understand your r(t), the third compoenent (the 90) is not a constant as the man is climbing.
 

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