A 35-N bucket of water is lifted vertically 3.0 m and then returned to its original position. How much work did gravity do on the bucket during this process?

0 J

Explanation:

From the question, let us assume that a positive work was done in lifting the bucket down and a negative work was done in lifting the bucket up.

Going up

W = -Fd Equation 1

Where W = work done,  F = Force needed to lift the bucket, d = distance

Given: F = 35 N, d = 3.0 m

Substitute into equation 1

F = -35(3.0)

F = -105 N.

Coming Down

W = Fd

Given: F = 35 N, d = 3.0 m

W = 35(3.0)

W = 105 N.

During the process (going up and coming down)

ΔW = -150+105

ΔW = 0 J

Hence the work done  by gravity during the process = 0 J

we need some answer choices, or something like that

explanation:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.