Problem :
What is the kinetic energy of a 2 kg ball that travels a distance of 50 meters
in 5 seconds?
The velocity of the ball is easily calculable: v = = 10 m/s. With
values for the mass and velocity of the ball, we can calculate kinetic energy:
K =
mv^{2} =
(2 kg)(10 m/s)
^{2} = 100 J
Problem :
In a sense we all have kinetic energy, even when we are standing still. The
earth, with a radius of 6.37×10^{6} meters, rotates about its axis once a
day. Ignoring the earth's rotation about the sun, what is the kinetic energy of
a 50 kg man standing on the surface of the earth?
To find the velocity of the man we must find how far he travels over a given
time period. In one day, or 86400 seconds, the man travels the circumference of
the earth, or 2Πr meters. Thus the velocity of the man is v = = = 463 m/s. Again, since we know the
velocity and mass of the man we can calculate kinetic energy.
K = mv^{2} = (50 kg)(463 m/s)^{2} = 5.36×10^{6}
Joules.
Problem :
A ball is dropped from a height of 10 m. What is its velocity when it hits the
ground?
The ball is acted upon by a constant gravitational force,
mg. The work done
during its total trip, then, is simply
mgh. By the WorkEnergy theorem, this
causes a change in kinetic energy. Since the ball initially has no velocity, we
can find the final velocity by the equation:
W = ΔK
mgh =
mv^{2}
Solving for v
,
v =
=
= 14 m/s
The final velocity of the ball is 14 m/s. We found this by one simple
calculation, avoiding the cumbersome kinematic equations. This is an excellent
demonstration of the advantages of working with the concepts of work and energy,
as opposed to simple kinematics.
Problem :
A ball is thrown vertically with a velocity of 25 m/s. How high does it go? What
is its velocity when it reaches a height of 25 m?
The ball reaches its maximum height when its velocity is reduced to zero. This
change in velocity is caused by the work done by gravitational force. We know
the change in velocity, and hence the change in kinetic energy of the ball, and
can calculate its maximum height from this:
W = ΔK
mgh =
mv_{f}^{2} 
mv_{o}^{2}
But v_{f} = 0
, and the masses cancel, so
h =
=
= 31.9 m
When the ball is at a height of 25 meters, the gravitational force has done an
amount of work on the ball equal to W =  mgh =  25 mg. This work causes a change
in velocity of the particle. We now use the WorkEnergy Theorem, and solve for
the final velocity:

mgh =
mv_{f}^{2} 
mv_{o}^{2}
Again, the masses cancel:
v_{f}^{2} =
v_{o}^{2} 
gh
Thus
V_{f} =
=
= 11.6 m/s
Problem :
A ball with enough speed can complete a vertical loop. With what speed must the
ball enter the loop to complete a 2 m loop? (Keep in mind that the velocity of
the ball is not constant throughout the loop).
At the top of the loop, the ball must have enough velocity such that the
centripetal force provided by its weight keeps the ball in circular motion. In
other words:
F_{G} =
F_{c} thus
mg =
Solving for v
,
v =
=
= 4.4 m/s
This value for the velocity gives us the minimum velocity at the top of the
loop. But we are asked for the minimum velocity at the
bottom of the
loop. How do we find this? You guessed it: WorkEnergy Theorem.
During the entire vertical loop, the ball is acted upon by two forces: the
normal force and the gravitational force. The normal force, by definition,
always points perpendicular to the circumference of the loop, and thus the
motion of the ball. Consequently, it cannot perform work on the ball. The
gravitational force, on the other hand, does perform work on the ball, according
the height it reaches. Since the radius of the circle is 2 m, the ball reaches a
height of 4 m, and experiences work from the gravitational force of  mgh =  2mg.
Remember the sign is negative because the force acts in a direction opposite the
motion of the ball. This work causes a change in velocity from the bottom of the
loop to the top of the loop, which can be calculated by the workenergy theorem:
W = ΔK
Thus

mgh =
mv_{f}^{2} 
mv_{o}^{2}
Canceling the mass and solving for v_{o}
,
v_{o} =
=
= 7.7 m/s
Thus the ball must enter the vertical loop of at least 7.7 m/s.