5 CONSERVATION OF ENERGY 5.8 Power
where v = dr/dt is the object’s instantaneous velocity. Note that power can be
positive or negative, depending on the relative directions of the vectors f and v.
If these two vectors are mutually perpendicular then the power is zero. For the
case of 1 - dimensional motion, the above expression reduces to
P = f v. (5.57)In other words, in 1 - dimension, power simply equals force times velocity.
Worked example 5.1: Bucket lifted from a well
Question: A man lifts a 30 kg bucket from a well whose depth is 150 m. 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 gravita-
tional force fJ acting on the bucket is of magnitude m g and is directed vertically
downwards. Hence, fJ = −m g (where upward is defined to be positive). The net
upward displacement of the bucket is h. Hence, the work W J 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:
WJ = fJ h = −m g h.Note that WJ 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 which balances (and
very slightly exceeds) the force due to gravity. Hence, f = −fJ. It follows that the
work W done by the man is
W = f h = m g h = 30 × 150 × 9.81 = 4.415 × 104 J.Note that the work is positive, which implies that the man expends energy whilst
lifting the bucket. Of course, since W = −W J, the energy expended by the man
equals the energy gained by the gravitational field.