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A question I found in a book recently concerned two worms climbing up one side of a wall and back down the other side, such that the worms were each half across the wall (I.e. the lengths on either side of the wall were equal for each worm). To calculate the amount of work done against gravity, it is suggested that this can be done by using the change in height of the worm's centre of gravity, in the equation W = mgh.

My question is: does this imply that negative work can be done against gravity?

My initial instinct would be yes, since the worm starts with a height of the centre of gravity of 0m, which is raised (implying +ve work against gravity) and then lowered again (implying -ve work against gravity).

The second question is: how come the snail doesn't do total work of: the work done to raise half of its mass by the height of the wall + the work done to raise half of its mass to mean height of the trailing half above the wall?

More generally, if I (70kg) climb a ladder of height 1m and then climb back down again, have I done 70g N work against gravity? Or have I done 0N work?

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  • $\begingroup$ Besides the answer by G. Smith, remember that in reality, besides the work "against" gravity a human would also have to do more work due to friction, not to mention the energy expenditure required to be alive. $\endgroup$
    – Ertxiem - reinstate Monica
    Commented Mar 26, 2019 at 1:13
  • $\begingroup$ Where did that snail come from? I thought we were discussing two worms. ;) $\endgroup$
    – PM 2Ring
    Commented Mar 26, 2019 at 2:05
  • $\begingroup$ Sorry, I was quite tired when I wrote this and I think it shows. One of the worms got fed up of being a worm and evolved into a snail ;) $\endgroup$
    – user19642323
    Commented Mar 26, 2019 at 7:13

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If you climb up and back down the same distance, the net work you have done against gravity is zero. Climbing up involves you doing positive work; climbing down involves you doing negative work.

You don’t have to come back down the same way you went up. Over any closed loop, the net work you will do is zero.

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  • $\begingroup$ Thank you, sorry for the confusing wording of my question. This is basically the answer I was looking for. $\endgroup$
    – user19642323
    Commented Mar 26, 2019 at 7:13
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does this imply that negative work can be done against gravity?

This isn't a very well-worded questions. Forces do work. This work is independent of what other forces may be acting on the object$^*$. So you should just ask "can this force do negative work?" Whether or not gravity is also acting on the object and whether or not these forces oppose each other is irrelevant. There is no "work against gravity".

More generally, if I (70kg) climb a ladder of height 1m and then climb back down again, have I done 70g N work against gravity? Or have I done 0N work?

Assuming a constant force each time, if you apply $700\ \rm N$ over $1\ \rm m$ in the same direction as your displacement then you do $700\ \rm J$ of work, and if you apply $700\ \rm N$ over $1\ \rm m$ in the opposite direction as your displacement then you do $-700\ \rm J$ of work. Notice how this doesn't depend on what gravity is doing.

$^*$ Of course the other forces as well as the one in question determines the trajectory of the object, but if you know the trajectory then you only need to focus on the force in question to determine the work done by it over the path the object takes.

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