Work and Energy

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Work and Energy

• Work, Kinetic Energy, Potential Energy, and
  Conservation of Energy

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Work Done by a Constant Force
Work is the product of the displacement and the
force component in the direction of the
displacement W F cosq d (a
scalar) Where the units of work are the Joule,
J, 1 J 1 N m
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Example of a constant force work calculation.
F 250 N at 300 below horizontal. D 6.65 m
parallel to ground. How much of F is parallel to
ground F cos (300) 250 cos(300) 216.5
N, So the work done by this force is, W F
cosq d 216.5 N x 6.65 m 1440 Nm 1440 J
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Work Done by a Variable Force

• A variable force is one that depends upon a
  variable. For example, the drag force that acts
  against moving objects depends upon the velocity
  of the object.
• An example of a variable force is the force it
  takes to extend a spring
• F - k x,
• Where k is the spring constant.

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Hookes Law
A linear restoring force. The spring stores
energy and can return it by doing work. The
amount of work, W, is W ½ k x2
F lkxl
F, N
X, m
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Example
lFl 60 N/m x, where x goes from 0 to 0.15 m.
How much work does this force do? What is k?
k 60 N/m What is the definition of
work done by a spring? W ½ kx2 ½ (60
N/m)(0.15m)2 0.675 J
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Work-Energy Theorem
Since F ma, and W Fcosq d, then work must
result in an objects velocity being
altered. Recall, v2 v02 2 a x , multiply
by m, m v2 mv02 2 ma x, divide by 2 ½ mv2
½ mv02 F x, identify Fx as W ½ mv2 ½
mv02 W, define K ½ mv2 So, K K0 W, or
DK W
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Example
How fast will a 200 g object be moving after
being pushed by a 75 N over a distance of 60
cm? F 75 N, v0 0 m/s, v ?, d 0.60 m, m
0.20 kg. K ? What is initial kinetic energy, K0
So, DK K W Fd 75 N x 0.60 m 45 J. If
K 45 J ½ mv2, the v2 2(45J)/(0.20kg)450
m2/s2, So v 21 m/s.
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Another Example
If it takes 5000 J to take a car from 10 kph to
20 kph, how much work will it take to go from 20
kph to 30 kph? Given W 5000 J, when v0 10
km/h and v 20 km/h. W ? When v0 20 km/h to
v 30 km/h? Equation W DK ½ mv2 ½ mv02
Solution 5000 J ½ m (20,000 m/3600 s)2 ½ m
(10,000 m/3600 s)2 Solve for mass and use in
second calculation W ½ m (30,000 m/3600 s)2
½ m (20,000 m/3600 s)2 W J
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Potential Energy

• Energy that is associated with position in a
  force field.
• Examples
• Gravitational Potential Energy
• Spring Potential Energy
• U, the symbol for potential energy.
• Units for PE are Joules, J

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Potential Energy-Zero Point
The amount of PE is calculated relative to a zero
point. For the spring the zero point is the
equilibrium length of the spring. U ½ kx2 ½
kx02 For gravity, the zero point is ground
level. U mgy mgy0
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Example
Yattic 4.5, y0 0, ybase -3.0, m 1.5 kg
4.5m
Ua 1.5 kg 9.8 m/s2 4.5 m
U01.5kg9.8m/s2 0m 0 J
0
Ub1.5kg9.8m/s2-3 m
-3.0 m
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Example
How much would you have to stretch a k 45 N/m
spring to store 1 J of energy? U 1 J, k 45
N/m, x ???? U ½ k x2 X 2 U/k1/2 2 ( 1
J)/ ( 45 N/m)1/2 0.21 m 21 cm
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Conservation of Energy
Total Mechanical Energy, E K U. In physical
phenomena, total mechanical energy is conserved,
Ef Eo Kf Uf Ko Uo
½ mvf2 Uf ½ mvo2 Uo
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Energy and Non-conservative Forces
Since experience and experiment have validated
the conservation of energy principle in all
cases, we know that when it appears to fail, it
only means that we have not accounted for all the
energy and work contributions.
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Continued

• What if we find,
• Ef E0 gt 0 ?
• Then, some of the energy that was available
  initially has not manifest itself in the final
  state.
• E0 Ko Uo and Ef Kf Uf , so
• Kf Uf (Ko Uo) gt0
• Kf - Ko Uf - Uo gt0

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Continued

• DK DU gt0
• This means that a non-conservative force has done
  work in the problem and consumed some of the
  energy.
• Wnc Example friction when working against
  friction, the path taken makes a difference.

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Power- the rate of doing work
P, symbol for power. P W/t, so units are J/s
Watt (W) Alternate expressions Since W Fd,
then P W/t Fd/t Fv Efficiency What you get
out versus what you put in e Wout/Win
Pout/Pin
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Problem
A 3250 kg plane reaches a cruising altitude of
10,000 m moving at 850 km/h in 12.5 minutes. If
the jet engines supply 1500 hp, what is the
efficiency of the plane? M 3250 kg, t 12.5
min 750 s, y 10,000 m V 850 km/h 236 m/s,
Pin 1500 hp 1119000 Watts.
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Continued
e P(out)/P(in) W(out)/t / 1119000 Watts e
(DKDU)/t/1119000 (½ mv2
mgy)/t/1119000 Watts (0.53250kg(236m/s)2
3250kg9.8m/s210000m)/750/1119000 Watts
(90506000318500000)750/1119000 0.48 or 48
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