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Physics 1D03 - Lecture 19
Work and Energy
•
•
•
•
Review of scalar product of vectors
Work by a constant force
Work by a varying force
Example: a spring
Physics 1D03 - Lecture 19
Work and Energy


F

m
a
Newton’s approach: 
- 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:

ˆ  Ay ˆj  Az kˆ
A

A
i
x

B  Bx iˆ  By ˆj  Bz kˆ
 
Then, A  B  Ax Bx  Ay By  Az Bz
(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 parallel unit vectors
Physics 1D03 - Lecture 19
Work

F

F


FII

d
Work by a constant force F during a displacement s:
Work = (component of F parallel to motion) x (distance)
We can be write this as:
Work = F • d
= Fdcos(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
Quiz

A constant force F  (1iˆ  2 ˆj  3kˆ ) N is applied to
an

 object while it undergoes a displacement
The work done by F is :
d  (2iˆ  2 ˆj  2kˆ) m.
a) (2iˆ  4 ˆj  6kˆ) J
b) - 12 J
c) 12 J
d) 12 N
e) - 12 N
Physics 1D03 - Lecture 19
Example
A person walks down the stairs. Is the work done by
gravity positive or negative ?
What if the person were to walk up the stairs ?
Physics 1D03 - Lecture 19
Example
You are pulling a sled with a force of 50N at an angle of 30o
to the horizontal, for a distance of 25m. Determine the
work done by you in pulling the sled.
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  102 J  32 J
c) 40 J
Physics 1D03 - Lecture 19
Example
Fp = 120
w = 100 N
fk = 50 N
n
A block is dragged 2.5 m up along the slope. Which
forces do positive work?
A) negative work?
B) zero work?
Physics 1D03 - Lecture 19
Example
Fp = 120
w = 100 N
fk = 50 N
n
The block is dragged 2.5 m up along the slope of 37o. Find :
a)
b)
c)
d)
e)
work done by Fp
work done by fk
work done by gravity
work done by normal force
Total work on the block
Physics 1D03 - Lecture 19
a)
Wp = (120 N)(2.5 m) = 300 J
b)
Wf = (- 50 N)(2.5 m) = -125 J
2.5m
d=2.5m sin(37o)
F=mg
c)
o
Wg =  (100 N)  (2.5 m)  sin (37 )   150 J
d)
Wn = 0
( motion)
Total : 300 + (- 125) + (- 150) = 25 J
Physics 1D03 - Lecture 19
5 min rest
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
Variable Force:
Determine the work done by a force as the particle
moves from x=0 to x=6m:
F(N)
5
x(m)
0 1
2
3 4 5 6
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
Dx
In the limit as Dx goes to zero,
F ( x) 
dW
dx
Physics 1D03 - Lecture 19
An Ideal Spring
Hooke’s Law: The tension in a spring is
proportional to the distance stretched.
or,
F = kx
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 and E=½kx2
Physics 1D03 - Lecture 19
Example:
Show that for a spring, E= ½ kx2.
Physics 1D03 - Lecture 19
Quiz
A spring is hanging vertically. A student
attaches a 0.100-kg mass to the end, and
releases it from rest. The mass falls 50
cm, stretching the spring, before stopping
and bouncing back.
During the 50-cm descent, the total work
done on the mass was:
a)
b)
c)
d)
zero
0.49 J
-0.49 J
none of the above
Physics 1D03 - Lecture 19
Quiz
A physicist uses a spring cannon to shoot a ball at a
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