Alan

Hmm.  I've re-read the question more carefully.  

It says to treat the runner as a point mass, m and so implicitly assumes that the force she applies to the mass comes from outside the mass.  ie her leg muscles are outside of herself, though their mass is included within!  In this case her (massless) muscles do indeed provide an upward force on her (massive) point body, and we can think of the force as being upward!

So, choose your preferred weirdness: pushing the Earth down or massless legs pushing a point body up!

I think the latter is a dangerous way of looking at this as it is may too easily lead to confusion in more complicated situations where there could be several internal forces that actually contribute nothing to the motion of the object in question (a point has no internal structure).

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The question is worded sloppily in my view.  The runner does work using a downward force to move the Earth down by a certain distance, so the two negatives cancel each other out and the runner does some positive work!

Of course, that's a slightly weird way of looking at the situation, but if you are going to insist on careful use of signs in calculating the work (as the question does) then that's the way it should be done.

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Resolve the (left-facing, horizontal) force parallel to the line L and perpendicular to it. Only the parallel component counts towards the work done.  The angle θ is the smaller angle between F and L so the component Fcosθ points along L in the direction from A to B.

This is the same direction in which it pointed in part A of the question.  However, the movement is now in the opposite direction from part A, so it's not surprising that the sign is just the opposite of that in part A.

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I don't think she is applying an upward force.  She is applying a downward force on the Earth (through the stairs) and so is moving the Earth (and hence the stairs) downward, relative to herself!

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Select the word "Scientific" in the input area, from the pop-up menu select "Equations" then choose "2 variables, 2 equations"

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If Jack can lift a mass of 62kg on Earth, then he is lifting a weight of 62*9.8 N (= 607.6N).

Gravitational acceleration on the alien planet is 9.8*62/22 m/s² (= 27.6 m/s²)

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I guess you mean that the weight is 80000N.

Pressure is force/unit area, so pressure on the ground = 80000/(4*0.25) N/m² or 80000 N/m²

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Since the component of force in the direction of movement is now Fcosθ, the work done is just FLcosθ

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Alternatively, if you prefer a numerical solution, continue using Heron's formula for the area and:

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Here's a graphical solution to save having to bother with all that messy stuff!

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