Work !
•
•
•
•
Review of scalar product of vectors
Work by a constant force
Work by a varying force
Example: a spring
Serway & Jewett 7.1 – 7.3
Physics 1D03 - Lecture 19
Work and Energy
Newton’s approach:



F  ma
- acceleration at any instant is caused by forces
Energy approach: Net work = increase in kinetic energy
- equivalent to Newton’s dynamics
- scalars, not vectors
- compares energies “before and after”
Physics 1D03 - Lecture 19

B
Math Review
The scalar product or dot product of
two vectors gives a scalar result:


A
vector • vector = scalar
 
A  B  (magnitude

of A ) x (component

of B parallel

to A )
 ABcos(  )
Physics 1D03 - Lecture 19
ScaIar product and cartesian components:

A  A x iˆ  A y ˆj  A z kˆ

B  B x iˆ  B y ˆj  B z kˆ
Then,
 
A  B  A x B x  Ay B y  Az B z
(note the right-hand-side is a single scalar)
 
To prove this, expand A  B using the laws of arithmetic
(distributive, commutative), and notice that
iˆ  ˆj  ˆj  kˆ  iˆ  kˆ  0
since, i, j, k are mutually perpendicular
and iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  1 since they are unit vectors
Physics 1D03 - Lecture 19
MATH QUIZ

F  (1iˆ  2 ˆj  3 kˆ ) N
A constant force
is applied to
an

 object while it undergoes a displacement
s  ( 2 iˆ  2 ˆj  2 kˆ ) m. The work done by F is :
a)
b)
( 2 iˆ  4 ˆj  6 kˆ ) J
2
2
2 4 6
2
J
c) 12 J
Physics 1D03 - Lecture 19
Work

F

F


FII

s
Work by a constant force F during a displacement s:
Work = (component of F parallel to motion) x (distance)
We can also write this as:
Work = F • s
= Fscos(θ)
Units : N • m = joule (J)
This is the “scalar product”, or “dot product”. Work is a scalar.
If work is done on a system, W is positive (eg: lifting an object).
If work is done by a system, W is negative (eg: object falling)
Physics 1D03 - Lecture 19
Example
(massless pulleys,
no friction)
FP
s=2m
100 N
How much work is done on the rope by Fp?
How much work done by the upward force on the ball?
Physics 1D03 - Lecture 19
Quiz
Fg = 5 N
P = 10 N
2m
3m
The two forces, P and Fg are constant as the block
moves up the ramp. The total work done by these
two forces combined is:
a) 20 J
b) 30 2  10 2 J  32 J
c) 40 J
Physics 1D03 - Lecture 19
Quiz
Fp = 120
w = 100 N
5
fk = 50 N
3
4
n
The block is dragged 2.5 m along the slope.
Which forces do positive work?
negative work?
zero work?
Physics 1D03 - Lecture 19
Quiz
Fp = 120
w = 100 N
5
fk = 50 N
3
4
n
The block is dragged 2.5 m along the slope. Find :
a) work done by Fp
b) work done by fk
c) work done by gravity
d) work done by normal force
e) Total work on the block
Physics 1D03 - Lecture 19
Forces which are not constant:
Example: How much work is done to stretch a spring scale from zero
to the 20-N mark (a distance of 10 cm)?
We can’t just multiply “force times distance” because the force
changes during the motion. Our definition of “work” is not complete.
Varying force: split displacement
into short segments over which F is
nearly constant.
F(x)
xi
xf
x
For each small displacement Dx, the
work done is approximately F(x) Dx,
which is the area of the rectangle.
F
Dx
Physics 1D03 - Lecture 19
We get the total work by adding up the work done in all the small steps. As we let
Dx become small, this becomes the area under the curve, and the sum becomes
an integral.
F(x)
F(x)
A
xi
xf
x
Split displacement into short
steps Dx over which F is nearly
constant...
W 
 F  Dx
xi
xf
x
Take the limit as Dx 0 and the
number of steps  
xf
W 
F
dx
xi
Work is the area
(A) under a graph of force vs. distance
Physics 1D03 - Lecture 19
xf
In 1D (motion along the x-axis):
W 
F
dx
xi
Another way to look at it: Suppose W(x) is the total work done in
moving a particle to position x. The extra work to move it an
additional small distance Dx is, approximately, DW  F(x) Dx.
Rearrange to get
F ( x) 
DW
In the limit as Dx goes to zero,
Dx
F (x) 
dW
dx
Physics 1D03 - Lecture 19
Example: An Ideal Spring.
Hooke’s Law: The tension in a spring is
proportional to the distance stretched.
or,
|F| = k|s|
The spring constant k has units of N/m
Directions: The force exerted by the spring when it is
stretched in the +x direction is opposite the direction of the
stretch (it is a restoring force):
F = -kx
Physics 1D03 - Lecture 19
Example: Work by a Spring
Fs
xi
Find a function W(x) so that
xf
Fs - kx
Fs ( x ) 
dW
dx
Physics 1D03 - Lecture 19
Quiz
A physicist uses a spring cannon to shoot a ball at a stuffed
gorilla. The cannon is loaded by compressing the spring 20 cm.
The first 10 cm of compression requires work W. The work
required for the next 10 cm (to increase the compression from 10
cm to 20 cm) would be
a)
b)
c)
d)
W
2W
3W
4W
Physics 1D03 - Lecture 19
Summary
For a constant force,
Work = F • s
= (component of force parallel to motion)  (distance)
For a non-constant force (1-D) : Work =  Fx(x) dx
Practice problems: Chapter 7, Problems 1, 7, 11, 13
(5th ed) Problems 5, 11, 15, 17
Read section 7.2 (scalar product).
Physics 1D03 - Lecture 19