Energy and Work

1
Energy and Work

• Force Times DistanceForce must be appliedObject
  must move
• Force must be the component which is in the same
  direction as the Distance
• This makes the formula
• W F x cos q

2
Energy and Work

• Both are proportionalTwice the force twice the
  workTwice the distance twice the workHolding
  heavy object no work Object not moving has no
  work on it
• Work is happening within muscles, not outside

3
Example

• A child pulls a toy 2.0 m across the floor by a
  string, applying a force of constant magnitude
  0.80 N. During the first meter the string is
  parallel to the floor. During the second meter
  the string makes an angle of 30o with the
  horizontal direction. What is the total work done
  by the child on the toy?
• Calculate the work separately for the first and
  second parts of the motion and then add them
  together. Since work is a scalar, add directly
  and do not use vector addition. For the first
  part, the work W is
• W F1x1 cos q where q 0o W1
  (0.80N)(1.0m)(1.0) 0.80 J
• For the second part of the motion, the work W2 is
• W2 F2x2 cos q 2, where q 2 30o W2 (0.80
  N)(1.0 m) (0.87) 0.69 J
• The total work W is then
• W W1 W2 0.80 J 0.69 J 1.5 J

4
Categories of work

• Against another forceArcher pulling bow
  stringPush-ups against gravityPushing object
  against friction
• To change speed of an object

5
Units of work/energy

• Joule - one Newton times one meter
• Calorie - 4.18 joules, used for heat

6
Work by a Varying Force

• If force is not constant, but changing, the
  equation must also change.
• Spring force depends on stretch of the spring by
  the formula F k x , where k is the spring
  constant for any particular spring

7
Work with Varying Force

• The area under the graph is equal to the work.
  This becomes
• 1/2 F x 1/2 (kx)x 1/2 kx2
• Work in a varying situation therefore will depend
  on the proportionality constant in that situation

8
Example
How much work is required to extend an exercise
spring by 15 cm if the spring constant is 800
N/m? W 1/2 kx2 1/2 (800)(.15)2 9.0 J
9
Homework!!

• P. 141ff 1,3,4,5 and internet question from
  drrileys site

10
Mechanical Energy

• Energy causes work
• Energy comes in many forms
• Energy changes between forms
• Two Basic Types
• Energy of movement - kinetic
• Energy of position - potential

11
Kinetic Energy

• When object is moving - energy is usedK.E. 1/2
  mass times velocity times velocity 1/2 mv2
• Moving objects can cause others to move, thus
  causing work to be done
• Velocity difference causes greater change in
  K.E.Double v quadruple K.E.Car going 100 km/h
  takes four times as long to stop as a car going
  50 km/h

12
Relating Work to KE

• If WFx and F ma then W max
• KE involves velocity, so using vf2v02 2ax ,
  solving for ax and inserting into the equation
• W 1/2 mvf2 - 1/2mv02
• But since KE 1/2 mv2 , this becomes
• W D KE 1/2 mvf2 - 1/2mv02
• which is called the work-energy theorem

13
Example
A physicist determines that an alpha particle had
an initial kinetic energy of 8.0 x 10-14 J The
mass of an alpha particle is known to be 6.65 x
10-27 kg. What was the initial speed of the alpha
particle in m/s? What was the speed when
expressed as a fraction of the speed of light (c
3 x 108 m/s)? Calculate the particle's speed
from the definition of kinetic energy. From the
definition of kinetic energy KE 1/2
mv2 Solve for v to give Inserting the
numerical values gives 4.9 x 106
m/s Comparing the speed of the alpha particle
to the speed of light C gives v 4.9 x 106
m/s 0.016, or v 0.016 c c
3 x 108 m/s
14
Example
(a) How much work is done to move a 1000-kg Mazda
MX-5 Miata automobile from rest to 27 m/s (60
mi/h) on a level road? (b) If this takes place
over a distance of 140 m, what is the average net
force? Use the work-energy theorem to find the
work. w 1/2 mv22 - 1/2 mv12 We set v1 equal to
zero, v2 27 m/s, and m 1000kg. Then the work
is W 1/2 (1000 kg)(27 m/s)2 3.6 x 105 J (b)
The average net force can be found from W F x, F
W 3.6 x 105J 2.6 x 103 N. x
140m
15
K.E. in many forms

• Heat - molecular motion
• Sound -molecular vibration
• Light - photon motion

16
Potential Energy

• Energy due to position
• Position can be due to Gravity or Electromagnetic
  forces
• Energy which is storedGravitational potential
  energy weight times heightWeight mass times
  gravity P.E. mgh
• P.E. same no matter how it gets to its position

17
Potential Energy
18
Example
A 500-kg mass pile driver is dropped from a
height of 3.00 m onto a piling in the ground. The
impact drives the piling 1.0 cm deeper into the
ground. If all the original potential energy of
the mass is converted into work in driving the
piling into the ground, what is the frictional
force acting on the piling? Assume the frictional
force to be constant over the 1.0-cm travel.
Solve this problem by using the concept of
energy. The driver falls through a distance h,
striking the piling and driving it into the
ground. Assume all of the potential energy goes
into causing this motion. In driving the piling,
the driver must overcome the frictional force
between the ground and the piling. The frictional
force will be opposite in direction to the force
exerted by the driver. The work needed to drive
the piling through a distance x is W F
x, where F is the force needed to overcome
friction. The initial potential energy relative
to the final position is PE mgh. The final
potential energy is 0. Equate the work done and
the change in potential energy to give Fx mgh
-0, F mgh (500 kg)(9.8 m/s)(3.00 m)
1.5 x 106 N. x
0.010 m
19
Homework!!

• P. 142ff
• 7, 9, 10, 15, 16

20
Conservation Law -

• Energy not created or destroyed can be
  transformed but total is constantPendulum shows
  this
• PE --gt KE --gt PE --gt KE --gt PE --gt etc.Even
  mass is considered concentrated energy, released
  by nuclear fission in nuclear reactors or by
  nuclear fusion in sun and stars
• Total Energy is constant
• ET KE PE

21
Conservation of energy

• Energy transformsSnow melts in mountain
  potential to kineticWater turns turbine in
  powerhouse kinetic straight to kinetic
  radialTurbine turns generator kinetic radial
  to electricityElectricity runs light in class
  electricity to lightLight causes calculator to
  work light to electricity Other pathways
  similar, but always there is less available
  energy after transformation
• Some becoming heatHeat is least useful form
  because it spreads most

22
Example
A student accidentally knocks a plant off a
window sill, where it falls from rest to the
ground 5.27 m below. Use the principle of
conservation of energy to determine its speed
just before it strikes the ground. The total
mechanical energy E of the plant is a constant.
It consists of two parts kinetic energy KE and
potential energy PE, which are not individually
constant, but whose sum is constant E KE
PE.
By expressing these energies in terms of the
height h and the final velocity v, determine v in
terms of h. Using subscripts T for top and 0 for
ground, equate the total energy at the top of the
fall to the total energy at the bottom,
ET E0 KET PET KE0
PE0. But KET 0 because the initial speed is
zero, and PE0 0 because h0. Thus we have
PET KE0 or mgh 1/2mv2. Upon solving for
v, we get v
10.2 m/s
23
Potential/Kinetic Interplay

• Consider a roller coaster track as shown

The height of the track above some reference
level determines the potential energy at that
point. In part (a) a car of mass m leaves point A
with no initial velocity (vA 0). We can
determine its speed at any other location B using
the principle of conservation of energy KEA
PEA KEB PEB 0 mghA 1/2mvB2
mghB 1/2mvB2 mghA - mghB, or v
24
Energy Conservation vs. Nonconservative Forces

• Mechanical energy isnt conserved in situations
  where friction changes some of it to heat, so it
  must be included in the equation
• E0 Ef Wfr Where E is the mechanical energy
  and W represents the work of friction
• E would contain both K and U components in both
  initial and final positions of a moving object.

25
Example

• A 55-kg carton of bananas with an initial speed
  of 0.45 m/s slides down a ramp inclined at an
  angle of 23o with the horizontal. If the
  coefficient of friction between the carton and
  the ramp is 0.24, how fast will the carton be
  moving after it has traveled a distance of 2.1 m
  down the ramp?
• Find the speed of the carton by applying the law
  of conservation of energy.
• Einitial Efinal Wfriction
  1/2mv02 mgh0 1/2 mvf2 mghf Ffrx
• The distance traveled by the carton is x 2.1 m.
  The difference in height from the initial
  position to the final position is h0 - hf x sin
  23o, and the frictional force is mmg cos 23o.
  Inserting these values into the last equation, we
  get
• 1/2 mvf2 1/2mvi2 mg (h0 - hf) - Ffrx 1/2
  mvf2 1/2mvi2 mgx sin 23o- m mgx cos 23o.
• We can multiply each term by 2/m to get
• vf2 v02 2gx (sin 23o - m cos 23o)
• Now we can insert the numerical values for vi,
  g, x, and m to get
• vf
• vf 2.7 m/s

26
Power

• How fast work is or can be done Power work
  divided by time
• Greater power means same work done in less amount
  of time
• Unit of power - watt one joule per second
  Electricity measured in KilowattsMotors rated
  in horsepower 750 watts

27
Example
Using a calibrated treadmill, a jogger measures
her energy output as 4.8 x 105 J for an hour's
run. What average output power did she develop?
From the definition of power as the rate of doing
work, we have P W 4.8 x 105 J
t 3600 s 1.33x102 w 130
w. A 70-kg person runs up a staircase 3.0 m
high in 3.5 s. How much power does he develop in
climbing the steps? If we assume that the work is
done at a constant rate, and that the work done
is the change in gravitational potential energy,
mgh, so the power is P W mgh (70
kg)(9.8 m/s)(3.0 m) 590 w
t t 3.5 s
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Homework!!

• P. 142ff
• 17, 21, 22, 26, 29, 33, 35, 37, 39, 43

29
Homework!!

• P. 144ff
• 49, 52, 56, 59, 61, 64, 66, 69, 70