Transcript 4.Work_Energy

Work
• Work
–Moving an object with a force that is in the direction of
the movement.
W = F ∙ d
•If F and displacement moved are in same direction, W is (+)
•If F and displacement moved are in opposite directions, W is (-)
• Only forces that are in line with the displacement do work.
–Positive Work will speed up an object (ex: Pushing).
–Negative Work will slow an object down (ex: Friction).
–Work can also be done to move an object against an
existing force (ex: Lifting against gravity).
–S.I. Units: 1 joule = (1 J) = 1N ∙ m = 1kg ∙ m2/s2
Example: Work and Luggage
1. Joe pushes a 20 kg piece of luggage with a
force of 5 N across the floor for 3 meters.
What work does Joe do on the luggage?
WF = F ∙ d = 5N ∙ 3 m = 15 J
2. If Joe is pushing this luggage at constant
velocity, then what is the work done by
friction on the luggage?
f = -F (since v is constant) so:
Wf = f ∙ d = (-5N)∙(3 m) = -15 J
•
What is the net work done on the luggage?
WNET = WF + Wf = 0 or
WNET = FNET = (F + f)∙d = (5N-5N)∙(3m) = 0
Pulling at an Angle
• Suppose Joe pulls with 5 N at an angle so that he
is pulling: (4 N east + 3 N up) for 3 meters
(east). What work does he do?
W = 4 N ∙ 3m = 12 J
East
5N
– Note: Only the component
of the force that is in the
direction of the motion is
responsible for the work.
Work, Mass, and Velocity
• How can we relate the work we do on an object
to its mass and resulting velocity?
• Assume a constant force causing a constant
acceleration.
• Use kinematics equations 1 and 2:
d = ½ at2
v = at
• WNET = FNET ∙ d
= ma ∙ ½ at2
= ½ m a2t2
= ½ m (at)2 = ½ m v2 (any comments?)
Power
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Power is the rate at which work is done.
P = W/Δt
Low Power: lifting or pushing an object slowly.
High Power: lifting or pushing an object quickly.
S.I. Unit: 1 watt = 1 (W) = 1 J/s = 1 joule/second
Ex: If Joe pushes the luggage with 5N through 3
meters in 2 seconds, what power is he exerting?
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PAV = W/ Δt = F ∙ d / Δt = (5N)(3m)/(2s) = 7.5 W
Notice: P = W/ Δt = F ∙ (d / Δt ) = F ∙ vAV
Since vAV = (3m/2s); PAV= (5N ) (1.5 m/s) = 7.5 W
(This is an average power)
Force Fields
• Force field: A region in space where an object
feels a force depending on its position in the
region.
• Examples:
– Gravitational Field: Fg=mg (at Earth’s surface)
– “Elastic” Field: a mass on a spring.
As a spring stretches more, mass feels more force
Energy
• Energy is a measure of an object’s ability to do
work.
• Two types of Energy:
– Kinetic Energy (KE): depends on an object’s mass and
speed: K.E. = ½ mv2
– Potential Energy (P.E.): depends on an object’s
position in a force field (such as Earth’s gravitational
field).
– Also, thermal energy, but that’s really the kinetic
energy of atoms and molecules moving around.
Potential Energy
• Gravitational Potential Energy:
(Important Formula)
P.E. = mgh ; h = height above ground
Q: What minimum work must I do to lift a mass
(m) a height (h)?
• Objects “FALL” to lower potential energy.
– A ball falls to lower positions.
– A mass on a spring moves so as to relieve the
stretch of a spring.
Work Changes Energy
• Work is a process by which an object’s
– P.E. is increased or decreased.
– K.E. is increased or decreased.
– P.E. is turned into K.E. or K.E. is turned into P.E
• And the Work done on the object equals its
change in energy..
• Just for perspective, what is another process
by which an object’s energy can be changed?
– Heat
Examples of Work Changing Energy
• Pushing an object over a distance w/o friction
– Positive Work increases the kinetic energy
• An object sliding to rest due to friction
– Negative work decreasing kinetic energy
• Lifting an object from the floor to a shelf.
– Positive work increasing potential energy.
• An object falling off a cliff
– Gravity does (+) work transforming P.E. into K.E.
• An object, thrown upwards, slowing as it rises
– Gravity does (-) work transforming K.E. into P.E.
Work and Energy
•
The work done on an object equals its change
in energy.
1. Example: Pushing an object: W = ΔK.E.
Joe pushes a 20 kg piece of luggage with a force of 5 N
across the floor for 3 meters. What is the speed of
the luggage if it is initially at rest? Neglect
friction.
W = F ∙ d = 5N ∙ 3 m = 15 J
ΔK.E. = ½ m (vf2 – vi2) = ½ m vf2 (since: vi = 0 )
So: W = ½ m vf2 = 15 J
vf2 =2 (15 J)/20 kg ; vf = 1.2 m/s
Work and Energy: Examples (cont’d)
2. Lifting an object against gravity: W = ΔP.E.
Joe lifts a 15 kg piece of luggage a height of 2
meters from the floor to a shelf. What is the
change in potential energy of the luggage?
• Work to overcome gravity to lift luggage a
height (h) :
W = mg × h = (15 kg) (10 m/s2)×(2 m) = 300 J.
Then:
ΔP.E. = W = 300J
Work and Energy: Examples (cont’d)
3. The 15kg luggage lifted by Joe in the last example,
falls off the 2m shelf. In doing so, the work due to
gravity turns the P.E. into K.E. So:
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ΔP.E. is (+) or (-) ?
ΔK.E. is (+) or (-)?
Notice: ΔP.E. + ΔK.E. = 0 (“Energy Conservation”)
From last example: P.E. of luggage on shelf = 300J.
With what K.E. does the luggage hit the floor?
With what speed does the luggage hit the floor?
Could you have found this speed without knowing
about energy?
Energy Conservation
• Energy can neither be created nor destroyed. It
can only be transformed from one form to
another.
• Examples:
– Falling luggage: P.E.  K.E.
– Swinging Pendulum: P.E.K.E.P.E. and back
– Mass going up and down on a spring.
• More Examples:
– If I push a block and increase its K.E., how is energy
conserved?
– If a sliding block comes to rest on a table due to
friction, how is energy conserved?
Steps to Solve Energy Conservation Problems
To conserve energy between two places A and B:
1. Draw a picture, labeling positions A and B. Write
down given information; identify unknown
quantity.
2. Write an expression for the total Energy at A
(ETOTA) = P.E.A+ K.E.A
3. Write an expression for the total Energy at B
(ETOTB) = P.E.B+ K.E.B
4. Set expression for (ETOTA) equal to expression for
(ETOTB)
5. Solve for the unknown quantity.
Example 1
Example 2