6. work, energy, and power. the dot product 3 where is the angle between the vectors and a and b...
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6. Work, Energy, and Power
The Dot Product
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The Dot Product
A
B
where is the angle between the vectors and A and B are their magnitudes.
cosA B AB
The dot product is the scalar
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The Dot Product
A few properties of the dot product:
2
( )
A A A
A B B A
A B C A B A C
cosA B AB
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The Dot Product
C A B
C C A A B B A B B A
The definition of the dot product is consistent with standard trigonometric relationships. For example:
2 2 2 2 cosC A B AB
Law of cosines
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The Dot Product
ˆˆ ˆx y zB B i B j B k
ˆˆ ˆ
x y zA A i A j A k
The definition
where
x x y y z zA B A B A B A B
cosA B AB
implies
Work and Kinetic Energy
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Energy Principles
netF am
So far we have solved motion problems by1. adding up all the forces to get the net force2. and applying Newton’s laws, e.g.,
2nd Law
Another way to solve such problems is to use an alternative form of Newton’s laws, based onenergy principles.
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Energy Principles
We are about to deduce an important energy principle from Newton’s laws.
However, today physicists view energy principles, such as the conservation of energy, asfundamental laws of Nature that are independentof the validity of Newton’s laws.
We start with the concept of kinetic energy.
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Kinetic Energy
netF m
m
dK
dt
v va
v
dtv
d
First, take the dot product of the 2nd law withthe velocity v
2
1net
B K
A K
dKF vdt dt
dt
Next, integrate both sideswith respect to time along a path from point A to point B
212K mv is the kinetic energywhere
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Kinetic Energy
When the right-hand side is integrated, we obtain the difference between the final andinitial kinetic energies, K2 and K1, respectively:
2 2
1 1net 2 1
B K K
A K K
dKF vdt dt dK K K
dt
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Work
The left-hand side
net
B
AF vdt
is called net work
can be rewritten as
net net
B B
A A
drF dt F dr
dt
net
B
AW F dr
The quantity
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The Work-Kinetic Energy Theorem
2 1W K K K
The net work, W, done by the net force on an object equals the change, K, in its kinetic energy.
Energy is measured in joules (J):J = N m
Work can be positive or negative. Kinetic energy is always positive.
Work-Kinetic Energy Theorem – Examples
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Example – Lifting a truck
A truck of mass 3000 kg is tobe loaded onto a ship usinga crane that exerts a force of 31 kN over a displacement of 2m.
Find the upward speed oftruck after its displacement.
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Example (2)
g app 2 1W W W K K
Two forces act on the truck:1. Gravity w2. Force of crane Fapp
Apply the work-kinetic energytheorem
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Example (3)
net ne
n t
t
e
B B
A AW F dr F dr
F r
Since the forces are constant overthe displacement, we can writethe work as
that is, as the dot product ofthe net force and the displacement.
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Example (4)
g
(3000 kg)(9.81 N
c
/
os
kg
180
1)(2 m)( )
58.9 kJ
oW w y mg y
Work done on truck by gravity
app app app cos
(31 kN)(2 m)( )
62.0 J
0
1
k
oW F y F y
Work done on truck by crane
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Example (5)
2 21 1app 2 12 2gW W mv mv
From the work-kinetic energy theorem
app2 22 1
2 2
2( )
2( 58,900 J 62,000 J)0
3000
1.45m
kg
2.09 m / ss /
gW Wv v
m
we obtain:
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Example – Compressed Spring
Find work done on blockfor a displacement, x = 5 cm
Find speed of blockat x = 0
iF xk
Hook’s Law
m = 4 kgk = 400 N/m
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Example (2)
Compute work done2 2
1 1
2
1
2 2 21 12 12 2
( )
| ( )
B x x
A x x
xx
W F dr kx dx k x dx
kx k x x
m = 4 kgk = 400 N/m
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Example (3)
2 212 12
2 212
( )
(400 N/m)((0m) ( 0.05
0.500
m )
J
)
W k x x
m = 4 kgk = 400 N/m
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Example (4)
Now apply work-kinetic energy theorem
2 21 12 2f iW mv mv 2 2
f i
Wv v
m →
vi initial speedvf final speed
m = 4 kgk = 400 N/m
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Example (5)
Speed at x = 0
2m 2(0.500 J
0.5)
0s
mkg
/4
sfv
Why did we ignore gravityand the normal force?
m = 4 kgk = 400 N/m
Power
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Power
Power is the rate at which work is done, orenergy produced, or used.
If the change in work is W, in time intervalt, then the average power is given by
WP
t
while the instantaneous power is
0lim
t
W dWP
t dt
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Power
The SI unit of power is the watt (W) namedafter the Scottish inventor James Watt.
W = J / s
Example: A 100 watt light bulb converts electrical energy to light and heat at the rateof 100 joules/s.
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Power
Given a force F and a small displacement dr the work done is dW F dr
therefore, the power can be written as
dW drP F F v
dt dt
that is, the dot product of the force and the velocity.
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Example – Bicycling
A cyclist who wants to move at velocity v while overcoming a force F must produce apower output of at least P = Fv. At 5 m/s against an air resistance of F = 30 N, P = 150 W.
However, even going up a gentle slope of 5o, an82 kg cyclist (+ bike) needs to output 500 W!
air( sin )P Fv F mg v
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Summary
The work-energy theorem relates the net work done on an object to the change in its kinetic energy: W = ∆K
Work done on an object by a force is
Power is rate at which work is done
B
AF dr