**Problem : **

An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip?

The work done by the elevator over the 100 meters is easily calculable:
*W* = *mgh* = (1000)(9.8)(100) = 9.8×10^{5}
Joules. The total time of the trip can be
calculated from the velocity of the elevator:
*t* = = = 25 s
. Thus the
average power is given by:
*P* = = = 3.9×10^{4}
Watts, or 39 kW.

**Problem : **

An object in free fall is said to have reached *terminal velocity* if the
air resistance becomes strong enough to counteract all gravitational
acceleration, causing the object to fall at a constant speed. The exact value of
the terminal velocity varies according to the shape of the object, but can be
estimated for many objects at 100 m/s. When a 10 kg object has reached terminal
velocity, how much power does the air resistance exert on the object?

To solve this problem we will use the equation
*P* = *Fv* cos*θ*
, Instead of the
usual power equation, as we are given the velocity of the object. We merely need
to calculate the force exerted on the object by the air resistance, and the
angle between the force and the velocity of the object. Since the object has
reached a constant speed, the net force on it must be zero. Since there are only
two forces acting on the object, gravity and air resistance, the air resistance
must be equal in magnitude and opposite in direction as the force of gravity.
Thus
*F*
_{a} = - *F*
_{G} = *mg* = 98
N, pointing upwards. Thus the force applied by air
resistance is antiparallel to the velocity of the object. Thus:

**Problem : **

*Calculus based problem* Derive, using the equation
*P* =
, an
expression for the power exerted by gravity on an object that is in free fall.

Our first step must be to generate an expression for work. We have already seen
that the work done by gravity after a distance
*h*
of free fall is equivalent to
*mgh*
. Can we take a time derivative of this expression? Of course: since
*h*
is
a measure of displacement, its derivative will simply give us the velocity of
the object:
= = *mgv*
. Thus, at any time during an
objectís free fall, the power exerted by gravity is given by
*mgv*
. Recall that
*P* = *Fv*
. If we check our derived answer against this equation we find that we are
correct.