Work Done by Person Carrying Groceries Up a Staircase

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Homework Statement


A spiral staircase in a building is in the shape of a helix of radius 5 meters. Between two floors of the building, the stairs make one full revolution and climb by 4 meters. A person carries a bag of grocers up two floors. The combined mass of the persona and groceries is 70 kg and the gravitational force is 70g downward. Calculate the amount of work done by the person against gravity

Homework Equations

The Attempt at a Solution


Is the answer just -mgh = -560g ? Don't we only have to take into account the vertical distance traveled, and the information about the spiral is irrelevant?
 
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Mr Davis 97 said:

Homework Statement


A spiral staircase in a building is in the shape of a helix of radius 5 meters. Between two floors of the building, the stairs make one full revolution and climb by 4 meters. A person carries a bag of grocers up two floors. The combined mass of the persona and groceries is 70 kg and the gravitational force is 70g downward. Calculate the amount of work done by the person against gravity

Homework Equations

The Attempt at a Solution


Is the answer just -mgh = -560g ? Don't we only have to take into account the vertical distance traveled, and the information about the spiral is irrelevant?
I think you have the right idea, but wouldn't it be 280 g (in nt-m)? The helical stairway doesn't enter into the calculation -- just the change in vertical position.
 
Mark44 said:
I think you have the right idea, but wouldn't it be 280 g (in nt-m)? The helical stairway doesn't enter into the calculation -- just the change in vertical position.
Yeah, I think you're right. Also, why would it positive as opposed to negative?
 
Mr Davis 97 said:
Yeah, I think you're right. Also, why would it positive as opposed to negative?
He's applying a force in the opposite direction of the gravitational force. In any case, it seems unusual that the work done would be a negative quantity, but then it's been a long time since I took physics.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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