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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.
High Potential Energy Low Potential Energy
• The potential energy acquired by an object equals the work done against gravity or other forces to place it in position.
• 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= vi
2+2aΔd
vi=0
vf2=2aΔd vf
2=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
• = (50kg)(9.8m/s2)(5m)
• = 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
• = 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.
• 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-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