Transcript Work/Energy PowerPoint


Work
Energy
Kinetic
Energy
Potential
Energy
Mechanical
Energy
Conservation
of Mechanical Energy
 The capacity to do work
 The energy transferred to an object
equals the work done on the object  ET = W
Energy is present in many forms
 Heat (thermal energy)
 Kinetic Energy = “Motion Energy”
 Potential Energy = “Stored Energy”


Mechanical energy – Sum of Potential and Kinetic
Nuclear
 Sound
 Electromagnetic energy
 Chemical energy
W  F  d cos
The parallel component does work
The (F cos ) component does work
NO work done if there:
1. Is no movement
2. The force if perpendicular to the displacement
In which photo(s) is WORK being done?
no
yes
yes
no
If a 10 Newton force displaces a 20 kg block 40 meters
calculate the work done on the block by the force.
F = 10N
W = F× d cos
= (10N)(40m)cos0
= 400J
0
The Normal force (FN )
and weight (mg) do no
work in this problem,
WHY?
Work = Force  Distance
W  F  d cos
• The component of force parallel to the displacement DOES WORK
• The perpendicular component DOES NO WORK
•  - Between the force and displacement
• Scalar
• Measured in Joules (J)
kg  m2
N m 
J
2
s
Positive Work - Force and displacement in same direction.
F
d
Negative Work - Force opposite the displacement.
F
d
Zero Work - Force is perpendicular to the displacement.
F
d
If the Force is in the Direction of Motion:
In this case,  = 0 o, so:
If the Force is in the Opposite Direction of Motion:
In this case,  = 180o, so:
If the Force is Perpendicular to the Direction of Motion:
In this case,  = 90 o, so:
If the Object Being Pushed Doesn't Move:
In this case, x = 0, so:
A 10 N force acts 300 above the horizontal and displaces an object 5
meters horizontally how much work is done?
F= 10 N
300
W = F×d cos
W = F d cos 
= (10N)(5m) cos 300
= 43.3 J
How much work does gravity do on a 70 kg
person who falls 100 meters in the free fall ride?
W  F  d cos 
 (mg )  d cos 
 (70kg )(9.8m / s 2 )(100m) cos 00
 68, 600 Joules

Kinetic energy (KE) is the energy of a moving object.
1 2
KE  mv
2

Energy associated with motion and mass.
 kg×(m/s)2
= kg× m 2 /s 2
= J ( Joules )
A 500 kg car is traveling at 10 m/s, 10 sec. later it is traveling
30 m/s. Calculate the following:
1. The initial kinetic energy
2. The final kinetic energy
3. The change in KE
1 2 1
KEi = mv = (500kg)(10m/s)2
2
2
= 25,000 J
1
1
2
KE f = mv f = (500kg)(30m/s)2
2
2
= 225,000 J
KE = KE f - KEi = 225,000J - 25,000J
= 200,000 J
Work Energy Theorem: The change in
kinetic energy equals the work done.
W = KE f - KEi
1
1
2
F  d = mv f - mvi 2
2
2
Work  KE
W  F d
The animation shows a block of mass and initial speed v sliding
across a floor that is not frictionless. A kinetic friction force fk
stops block during displacement d. Thus we can relate work done
by friction to the change E in the system's energy
W f  KE
1
 fd  0  mvi 2
2
Moving hammer can do work on nail!
For hammer:
WH  KE   Fd
For nail:
WN  KE  Fd
Matt’s little red wagon with a mass of 4.6 kg moves in a straight line
on a frictionless horizontal surface. It has an initial speed of 10 m/s
and is pulled by Matt 4.0 m with a force of 18N in the direction of
the initial velocity. Use the work-energy relation
(WNet = KE) to calculate the wagons:
a. Change in Kinetic Energy
b. Final speed.
In a test of old sports car, it’s found that engines provided around
1,000 N of force. If the typical mass is 400 kg and they accelerate
from rest, how fast will they be going 100 m down the road?
WN = KE
Fd = KE
1
(1000N)(100m)= (100kg)(v f 2 - 0)
2
v f = 22.4m/s
How much work is done
holding a box in place on
an incline?
How much work is done pushing
a 15 kg box up a 30° incline at a
constant speed for 3 m
W  F  d cos 
 mg sin   d cos(0 )
0
Force vs. Distance
Area under curve equals the work done
Force vs. Distance
Area under curve equals the work done
Area above the curve – work is positive
Area below the curve – work is negative
Work done by a variable force
equals
Area under the curve
Gravitational Potential Energy (PE): The energy an object
has due to its height above a reference point.
PE  mgh
The potential energy change is independent of the path between
the initial and final points.
 kg× m/s  m
2
2
= kg× m /s
2
= J ( Joules )
link
PE=mgh
Can gains
Potential Energy
equal to mgh
One serving of Bagel Crisps
contains 543 kJ. How many pullups are needed to burn it off?
M = 60 kg
∆h = .5 m
 ∆PE = mg ∆H
 543000J = (60kg)(9.81N/kg)(.5m)(n)
 1845 = n
 But the human body is only abour 20%
efficient so, n is only 369!

Who does more work in lifting the respective equal
masses to the top of the incline at a constant speed?
Rufus the 5 kg cat falls 10 meters from above
the surface of the earth. Calculate:
1. His initial potential energy with respect to the ground
PEi  mgh  (5kg )(10m / s )(10m)  500 J
2
2. His potential energy 1 seconds after being released
d  vi t  12 at 2  12 (10m / s 2 )(1s )  5m
PE f  mgh  (5kg )(10m / s 2 )(5m)  250 J
3. His change in potential energy
PE  PEf  PEi  (250  500) J  250 J
4. Where did its energy go when it hits the ground?
Conservative Force:
The work done is independent of
the path taken. Only depends on the initial and final position. Ex:
Gravity
Non Conservative Force:
the path taken. Ex: Friction
The work done depends on
Don’t disregard the Conservation Laws