178 Chapter 4 Energy
Laboratory
The Relationship Between Work and Kinetic Energy
Let’s examine how we can derive an equation that expresses the relation-
ship between work and kinetic energy. Suppose we continue to impose
force F on a cart in motion, in a direction parallel to that cart’s velocity.
That cart has mass m and starts with an initial, uniform velocity of v.
That means an additional force is imposed on the object in motion.
At this time, the following is true:
work done on the object = Fd
Also, since we’ve represented the final velocity as v
2
, we can represent the
change in the object’s kinetic energy as the following:
change in kinetic energy = ½ mv
2
2
− ½ mv
1
2
And since we already know that the change in kinetic energy is equal to
the work done on the object, we can express the following relationship:
½ mv
2
2
− ½ mv
1
2
= Fd
Aha.
Initial velocity v
1
Final velocity v
2
Distance d, the distance that a force is applied
Force F
The Relationship Betwn Work and Kinetic Energy 179
We can also derive this equation another way. Since F is defined as con-
stant, the cart is experiencing uniform acceleration. Therefore, if we rep-
resent the cart’s acceleration with a, we know that the following must be
true:
v
2
2
v
1
2
(Why is this so? See expression w on page 85.) To get closer to our original
expression, we’ll substitute using Newtons second law:
F = ma, or rearranged just a little, a =
F
m
And we’ll get the following:
v
2
2
v
1
2
=
2Fd
m
Then if you simply multiply both sides by ½, you’re there!
½ mv
2
2
− ½ mv
1
2
= Fd
I can get it right if I calculate very carefully.
Braking Distance
andSpeed
Using what we
relationship betwn
kinetic energy and
work, let’s consider
a car’s braking
distance.
What do you
mean, exactly?
We, I gue it’s
not just for cars.
It’s the distance that
any object in motion
requires to stop,
given a certain
force in the
oosite direction.
Given that we know a
change in kinetic energy
is equal to the work
performed, we know that
the foowing must be true
of bringing an object in
motion to rest:
½ ma × spd
2
= force of the brakes ×
distance the brakes are alied
½ mv
2
= F
brakes
× d
If we reaange
the equation, we
can solve for the
braking distance!
d =
½ mv
2
F
brakes
This equation means that
the greater the ma
(m) and the spd (v) of
the vehicle become, the
greater the required
distance to brake (d).
And the larger the
force of the brakes
( F
brakes
), the shorter the
distance required to
come to a complete
stop.
But we’ve
multiplied
the spd by
itself!?
That means that the
braking distance (d ) is
proportional to the
spd raised to the
second power.
The braking
distance is
proportional to
the spd raised
to the second
power.
When the initial
spd is doubled...
does that mean the
braking distance is
Spd
braking distance
Aha, that is exceent
insight into their
relationship. It is
dangerous to aume
that the braking
distance is linearly
proportional to
acar’s spd.
Yeah,
definitely.
The braking distance
doubled.
For a bicycle, it’s
not such a big deal,
but for automobiles,
it can have serious
consequences.
For example, suose a
car is traveling at 40 km/h,
and its braking distance
is 10 m. If this same caris
traveling at 120 km/h,
or at a velocity thr
times higher, what is the
brakingdistance?
Um...
since the spd is
thr times higher, we
just have to square
that. So 3 × 3 = 9 times
greater, or
10 m × 9 = 90 m.

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