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