6.13 - Interactive checkpoint: an elevator
An elevator with a mass of 515 kg is
being pulled up a shaft at constant
velocity. It takes the elevator 3.00
seconds to travel from floor two to
floor three, a distance of 4.50 m.
What is the average power of the
elevator motor during this time?
Neglect friction.
Answer:P = W
6.14 - Work and energy
We have discussed work on a particle increasing its KE, and work on a system
increasing its PE. Now we discuss what happens when work increases both forms of
mechanical energy.
Because we are considering only KE and PE in this chapter, we can say the net work
done on an object equals the change in the sum of its KE and PE. Positive work done
on an object increases its energy; negative work decreases its energy.
Let’s also consider what happens when an object does work, and how that affects the
object’s energy. Consider a soccer ball slamming into the hands of a goalie. The ball is
doing work, forcing the goalie’s hands backwards. The ball slows down; its energy
decreases. Work done by an object decreases its energy. At the same time, this work
on the goalie increases her energy. Work has transferred energy from one system (the
ball) to another (the goalie).We will use the scenario in Example 1 to show how both an object’s KE and PE can
change when work is done on it. A cannon shoots a 3.20 kg cannonball straight up. The
barrel of the cannon is 2.00 m long, and it exerts an average force of 6,250 N while the
cannonball is in the cannon. We will ignore air resistance. Can we determine the
cannonball’s velocity when it has traveled 125 meters upward?
As you may suspect, the answer is “yes”.
The cannon does 12,500 J of work on the cannonball, the product of the force (6,250 N)
and the displacement (2.00 m). (We assume the cannon does no work on the
cannonball after it leaves the cannon.)At a height of 125 meters, the cannonball’s increase in PE equals mgǻh, or 3,920 J.
Since a total of 12,500 J of work was done on the ball, the rest of the work must have
gone into raising the cannonball’s KE: The change in KE is 8,580 J.
Applying the definition of kinetic energy, we determine that its velocity at 125 m is
73.2 m/s. We could further analyze the cannonball’s trip if we were so inclined. At the
peak of its trip, all of its energy is potential since its velocity (and KE) there are zero.
The PE at the top is 12,500 J. Again applying the formula mgǻh , we can determine
that its peak height above the cannon is about 399 m.Work and energy
Work on system equals its change in
total energyThe cannon supplies 6,250 N of
force along its 2.00 m barrel.
How much work does the
cannon do on the cannonball?
W = (F cos ș)ǻx = Fǻx
W = (6250 N)(2.00 m) = 12,500 J
(^134) Copyright 2007 Kinetic Books Co. Chapter 06