What follows is an algebraic justification for the method(s) for calculating work...

An object of mass 2 kg is at rest, so its kinetic energy is 0 Joules. Suppose a (net) force of 10 Newtons acts on it.

According to Newton's Second
Law, the object's
acceleration (=
F_{net}/mass) = 10 N/ 2 kg = 5 m/s^{2}, which, of
course, means that the
velocity of the object will
change by 5 m/s each second that the force acts.

In 2 seconds, its velocity will increase by (5 m/s^{2})(2
s) = 10 m/s. At this time, the object's kinetic energy (=
mv^{2}/2) = (2 kg)(10 m/s)^{2}/2 = 100 Joules.

Therefore, during this episode, the object's kinetic energy increased by 100 Joules (from 0 Joules to 100 Joules). According to the Work/Energy Equation then, 100 Joules of work must have been done on it.

On the other hand, the object's average velocity during this 2
seconds (= (v + v_{0})/2) = (10 m/s + 0 m/s)/2 = 5 m/s. The
object had an average velocity of 5 m/s for 2 seconds, so it traveled
10 meters. Notice that F_{net}^{.}distance = (10
N)(10 m) = 100 Joules.

Is it an odd coincidence that the work done on the object in this case = force x distance? No.

If the direction of the force is the same as the direction of the object's motion, work done by the force = force x distance.

Suppose an object of mass 2 kg is moving along at a
velocity of 10 m/s. Its kinetic energy ( = mv^{2}/2) = (2
kg)(10 m/s)^{2} = 100 Joules. Suddenly, a net force of 10
Newtons in a direction opposite to the velocity begins to act on the
object. Newton's Second Law says that the acceleration of the object
will be (-10 N)/(2 kg) = -5 m/s^{2}. The negative sign for
the force and acceleration indicate that they are in the direction
opposite to the velocity. In 2 seconds, the velocity of the object
will have decreased by (-5 m/s^{2})(2 sec) = -10 m/s - the
object will be stopped, and its kinetic energy will be 0 Joules.

During this exciting episode, then, the kinetic energy of the object changed by -100 Joules (from 100 Joules to 0 Joules.) According to the Work/Energy Equation, -100 Joules of work must have been done on it.

On the other hand, during this 2 second period, the object's
average velocity (= (v + v_{o})/2) was (0 m/s + 10 m/s)/2 = 5
m/s. Since the object had an average velocity of 5 m/s for 2 seconds,
it traveled 10 meters. Notice that force x distance = (-10 N)(10 m) =
-100 Joules. Here the work done is force x distance if you consider
the force to be negative. Otherwise:

If the direction of the force is opposite to the direction of the object's motion, the work done by the force = - force x distance.

Last update November 21, 2007 by JL Stanbrough