The document describes a worked example problem of calculating the work done to lift a 30kg bucket 150m from a well. It provides the steps of identifying the gravitational force as mg directed downward, the upward displacement as the depth h, and calculates the work done by gravity as the negative of the force times displacement, -mgh. It then notes that the man lifting must exert an equal and opposite force f to balance gravity, so the work done by the man is calculated as mgh, equaling 4.415 x 104 Joules.

1. Education

2. Worked Example
Bucket lifted from a well
Question:
A man lifts a 30kg bucket from a well whose depth is 150m. Assuming that the man lifts the
bucket at a constant rate, how much work does he perform ?
Answer:
Let m be the mass of the bucket and h the depth of the well. The gravitational force f ’
acting on the bucket is of magnitude mg and is directed vertically downwards. Hence,
f ’ = - mg. The net upward displacement of the bucket is h. Hence, the work W ’ performed
by the gravitational force is the product of the (constant) force and the displacement of the
bucket along the line of action of that force:
W ’ = f ’ h = - mgh
Note that W ’ is negative, which implies that the gravitational field surrounding the bucket
gains energy as the bucket is lifted. In order to lift the bucket at a constant rate, the man
must exert a force f on the bucket that balances the force due to gravity. Hence, f = - f ’ .
It follows that the work done by the man is :
W = fh = mgh = 30 x 150 x 9.81 = 4.415 x 104J

3. “Education is not the filling of a
bucket, but the lighting of a fire.”

4. Academic Education
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• Reading about negotiation is a waste of time
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• Peer debriefing