physics 1d03 - lecture 19 review of scalar product of vectors work by a constant force work by a...
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Physics 1D03 - Lecture 19 The scalar product or dot product of two vectors gives a scalar result: vector vector = scalar Math ReviewTRANSCRIPT
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Physics 1D03 - Lecture 19
• Review of scalar product of vectors• Work by a constant force• Work by a varying force• Example: a spring
Work and Energy
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Physics 1D03 - Lecture 19
Work and EnergyWork and Energy
amF
Energy approach: Net work = increase in kinetic energy
- acceleration at any instant is caused by forces
- equivalent to Newton’s dynamics- scalars, not vectors - compares energies “before and after”
Newton’s approach:
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Physics 1D03 - Lecture 19
)ABcos( )A toparallel of (component x )A of (magnitude
BBA
B
A
The scalar product or dot product of two vectors gives a scalar result:
vector • vector = scalar
Math Review
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Physics 1D03 - Lecture 19
ScaIar product and cartesian components:
kBjBiBBkAjAiAA
zyx
zyxˆˆˆˆˆˆ
zzyyxx BABABABA
Then,
(note the right-hand-side is a single scalar)
To prove this, expand using the laws of arithmetic (distributive, commutative), and notice that
BA
since, i, j, k are mutually perpendicular
since they are unit vectors1ˆˆˆˆˆˆ0ˆˆˆˆˆˆ
kkjjii
kikjji
and
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Physics 1D03 - Lecture 19
MATH QUIZ
A constant force is applied to an object while it undergoes a displacement The work done by is :
N )ˆ3ˆ2ˆ1( kjiF
m. )ˆ2ˆ2ˆ2( kjis F
mN 12 d)J 12 c)
J 12- b)J )ˆ6ˆ4ˆ2( a)
kji
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Physics 1D03 - Lecture 19
Work
Work by a constant force F during a displacement s:
s
IIF
F
F
Units : N • m = joule (J)
Work = (component of F parallel to motion) x (distance)
We can also write this as:
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)
Work = F • s = Fscos(θ)
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Physics 1D03 - Lecture 19
Example
(massless pulleys, no friction)
s = 2 m
How much work is done on the rope by Fp?
How much work is done by the upward force on the ball?
100 N
FP
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Physics 1D03 - Lecture 19
Quiz
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:
3 m
2 m
a) 20 Jb) c) 40 J
J 32 J 1030 22
Fg = 5 N
P = 10 N
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Physics 1D03 - Lecture 19
Quiz
5
4
3
2.5 m
fk = 50 N
w = 100 N
n
Fp = 120
The block is dragged 2.5 m along the slope. Which forces do positive work?
negative work?
zero work?
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Physics 1D03 - Lecture 19
Quiz
5
4
3
2.5 m
fk = 50 N
w = 100 N
n
Fp = 120
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 gravityd) work done by normal forcee) Total work on the block
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Physics 1D03 - Lecture 19
a) Wp = (120 N)(2.5 m) = 300 J
b) Wf = (- 50 N)(2.5 m) = -125 J
c) Wg =
d) Wn = 0 ( motion)
J 150 m 2.5N) 100(53
Total : 300 + (- 125) + (- 150) = 25 J
mg
IIFmg 53 s
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Physics 1D03 - Lecture 19
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.
Forces which are not constant:
ix fx
Varying force: split displacement into short segments over which F is nearly constant.
F(x)
x
FFor each small displacement x, the work done is approximately F(x) x, which is the area of the rectangle.
x
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Physics 1D03 - Lecture 19
Work is the area (A) under a graph of force vs. distance
dxFWx
x
f
i
ix fx
Split displacement into short steps x over which F is nearly constant...
F(x)
xix fx
F(x)
x
Take the limit as x 0 and the number of steps
xFW
We get the total work by adding up the work done in all the small steps. As we let x become small, this becomes the area under the curve, and the sum becomes an integral.
A
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Physics 1D03 - Lecture 19
In 1D (motion along the x-axis): dxFWx
x
f
i
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 x is, approximately, W F(x) x.
Rearrange to getx
WxF)(
In the limit as x goes to zero, dx
dWxF )(
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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
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Physics 1D03 - Lecture 19
ix fx
Example: Work by a Spring
Fs
Fskx
Find a function W(x) so thatdx
dWxFs )(
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Physics 1D03 - Lecture 19
Concept 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) Wb) 2Wc) 3Wd) 4W