PHYS 101 Lecture 10 - Work and kinetic energy
10 - 1
Lecture 10 - Work and Kinetic Energy
What’s important:
• impulse, work, kinetic energy, potential energy
Demonstrations:
• block on plane
• balloon with propellor
• conversion of work to P.E.
Work
In previous lectures, we investigated the effect of a force acting over a period of time.
Newton’s Second Law is sometimes written in the form
F = ma = m Δv / Δt
---->
F Δt = m Δv = Δ(mv) = Δp
impulse
In other words, a force acting through time gives the impulse.
What about a force acting through a distance?
F Δx ≡ Work ≡ W
(measured in Joules)
Suppose the force produces a constant acceleration a. Then
$ v 2 # v 2i '
$ v 2 # v 2i ' 1
1
2
2
W ! F • "x = F & f
= ma & f
= mvf # mvi
% 2a (
% 2a ( 2
2
At this point, work begins to look like an interesting quantity because it depends only on
the end - points (not on the path). Is it true in general? Yes, but only in the absence of
friction (proof requires integration, can be found in PHYS 120 notes).
We see that the work changes the kinetic energy 1/2 mv2 of the particle. In words,
The work done by an unbalanced force is equal to the change in the kinetic
energy of the object.
As a problem-solving technique, construct a free-body diagram to determine the
unbalanced force.
In three dimensions, we must generalize F Δx. Since force and displacement are both
vectors, then we expect
W ≡ F • Δr.
© 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.
PHYS 101 Lecture 10 - Work and kinetic energy
10 - 2
Example
A molecular motor can walk along a filament in a cell (actin or microtubule)
carrying a cargo, or contracting a muscle:
molecular motor
The force it exerts is about 5 pN, and the power stroke is 5 nm. What is the work done
in the power stroke?
W = F • Δx = 5 x 10-12 x 5 x 10-9 = 2.5 x 10-20 Joules.
This is about half the energy released during the hydrolysis of ATP.
Example
Find the work done by the centripetal acceleration for an object executing
uniform circular motion.
Δx
F
F = m a = -(mv2/r2) r
---> F ⊥ Δx
---> Work = 0
It’s no surprise that there is no work done, since there is no change in kinetic energy.
When the forces are not constant, we must find the area under the force vs. distance
curve:
F
W
x
We'll do an example of this for a spring in the following lecture.
© 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.
PHYS 101 Lecture 10 - Work and kinetic energy
10 - 3
Potential Energy
Consider the situation of an object sliding down a frictionless plane at an angle θ.
!
L
h
The force due to gravity is mg, and it does work on the block
W=F•L
= mg • L sinθ
= mgh.
(using h = L sinθ)
In turn, the work W results in a change in kinetic energy of the block:
mvbottom2 /2 - mvtop2 /2 = mgh
Suppose we now raise the block very, very slowly back up to its original position, by
applying a force which just balances the force due to gravity.
N
Fexternal
mg
The total work is zero, since the forces cancel out in the direction of motion. Similarly,
ΔK = 0 since the object is at rest at the top and bottom. The work we have done on
the block has not resulted in a change in kinetic energy!
Even though this is a nice consistent picture, we find it unsatisfactory because we are
doing work on the object. In order to differentiate what we do as external agents, and
what happens to the system in response to our work, we introduce the idea of potential
energy. Here
Wdone on the system = increase in potential energy U of the system
In some sense, the work we have done is "stored" as potential energy, free to be
converted into kinetic energy at some later time. The gravitational potential energy is
then U = mgh. We can now write down a conservation law which states that
© 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.
PHYS 101 Lecture 10 - Work and kinetic energy
!E = 0
where
E = U + K
10 - 4
(conservation of energy)
U has replaced W
or
In the absence of dissipative forces [heat, friction...] and of forces for
which it is not possible to define a potential energy, the total mechanical
energy of a system is constant. [dissipative forces = friction, viscosity, etc.]
Caveats:
1. Don’t double-count work and potential energy – use only one in a given problem.
2. Beware of the signs of work:
Wdone on the system raises its potential energy
Wdone by the system lowers its potential energy.
For a particular situation, Wdone on the system = - Wdone by the system (somewhat like subtrction is
the addition of a negative number).
3. The zero of the potential energy is arbitrary. Work only tells us the difference in
potential, never its absolute magnitude.
Example
The work required to separate two bodies attracted by gravity from r = R
to r = ∞, where r is the distance of separation, is
ΔU = Gm1m2 / R.
This work raises the gravitational potential energy of the objects. If we want to say
U(r = ∞) = 0 [i.e. objects at infinite separation have no gravitational PE], then we set
U = -Gm1m2 / R.
The - sign should not bother us. The “zero” of the potential energy is not defined
because we can only evaluate the difference in potential energy arising from the work
Ugravity
r
We can relate this expression for the gravitational potential energy to the more familiar
mgh by comparing U at the surface of the Earth Re with U at a distance Re+h: The two
masses involved are the mass of the object m, and the mass of the Earth Me.
First, we look at the force of gravity on the surface of the Earth, r = Re. According to
© 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.
PHYS 101 Lecture 10 - Work and kinetic energy
10 - 5
Newton's universal law of gravity,
G Mem !# G Me $&
F =
=
m = gm
Re2
" Re2 %
That is, in terms of the Earth's size and mass,
G Me
.
g=
Re2
Now, let's calculate the potential energy difference:
! U = [U at Re + h] " [U at Re ]
="
=
G Mem G Me m
+
Re + h
Re
G Me m #%
Re &(
1"
Re $ Re + h '
The term in the bracket can be approximated by
Re
R + h ! Re
h
h
.
1!
= e
=
"
Re + h
Re + h
Re + h Re
Then, the expression for ΔU becomes:
G Mem h
G Me
!U "
•
=
mh = gmh
Re
Re
Re2
© 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.