Transcript ENERGY

ENERGY
• Energy is that property whose possession enables something
to perform work.
• 3-types 1- Kinetic Energy- which is the energy
something possesses by virtue of its
motion.
2- Potential Energy- Which is the energy
something possesses by virtue of its
position.
3- Rest Energy- which is the energy
something possesses by virtue of its mass.
E= mc2.
• The potential energy acquired by an object equals
the work done against gravity or other forces to
place it in position.
High Potential Energy
Low Potential Energy
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P.E. = Work = FΔd = mgΔh
P.E. = mgΔh
Units = kg-m/s2•m = kg-m2/s2 = N-m = Joule (J)
The gravitational potential energy defined by
P.E. = mgh is expressed in relation to an arbitrary
reference level where h=0
eg. Sea level, street level, ground level, or floor
level are useful reference levels.
• F=ma,
a=F/m
• vf2= vi2+2aΔd
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vi=0
vf2=2aΔd  vf2=2(F/m)Δd
FΔd=1/2mvf2
Work done on the ball =K.E. of the ball
Fd = 1/2mv2
K.E.=1/2mv2
units= Joules (J)
* You can set energies equal to each other.
Work = P.E.
Work = K.E.
P.E. = K.E.
• If a 50-kg mass of steel is raised 5-m. What is its
potential energy?
• P.E. = mgh
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= (50kg)(9.8m/s2)(5m)
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= 2.5 x 103J
• If a baseball has a mass of 0.14-kg and is thrown
with a velocity of 7.5-m/s, what is its K.E.?
• K.E. = 1/2mv2
• K.E. = 1/2(0.14-kg)(7.5-m/s)2
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= 3.9J
• A 600-g hammer head strikes a nail at a speed of
4-m/s and drives it 5-mm into a wooden board.
What is the average force on the nail? In
Newtons and in lbs.
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Work=K.E.
Fd=1/2mv2
F=mv2/2d = (.6-kg)(4m/s)2/2(.005-m)
F=960-N
960N•.225lbs/1N=216lbs.
• Find the K.E. of a 1200-kg car when it is moving at 25•
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km/hr and when it is moving at 100-km/hr.
How much more energy does the car have when it is
moving at 100-km/hr?
v1=25-km/hr=7-m/s, v2= 100-km/hr=27.8-m/s
K.E.1=1/2mv2=(.5)(1200kg)(7m/s)2
=28,935J
K.E.2=1/2mv2=(.5)(1200kg)(27.8m/s)2
=463,704J
The 100-km/hr car has 16 times as much K.E. as it does
at 25-km/hr. Thus the 100km/hr car can do more work
and at high speeds can cause severe auto accidents.
• If Meagan pushes on a lawn
mower with a constant force
of 90.0-N at an angle of 40°
to the horizontal, how much
work does she do in pushing
it a horizontal distance of
7.5m?
• W=(Fcos40°)d
• W=(90.0N)(cos40°)(7.5M)
• W=517J