2. 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= mc 2.

3. The potential energy acquired by an object equals the work done against gravity or other forces to place it in position.

4. P.E. = Work = FΔd = mgΔh P.E. = mgΔh Units = kg-m/s 2 m = kg-m 2 /s 2 = 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.

6. F=ma, a=F/m v f 2 = v i 2 +2aΔd v i =0 v f 2 =2aΔd  v f 2 =2(F/m)Δd FΔd=1/2mv f 2 Work done on the ball =K.E. of the ball Fd = 1/2mv 2 K.E.=1/2mv 2 units= Joules (J) * You can set energies equal to each other. Work = P.E. Work = K.E. P.E. = K.E.

11. If a 50-kg mass of steel is raised 5-m. What is its potential energy? P.E. = mgh = (50kg)(9.8m/s 2 )(5m) = 2.5 x 10 3 J

12. 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. = 1/2mv 2 K.E. = 1/2(0.14-kg)(7.5-m/s) 2 = 3.9J

13. 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=1/2mv 2 F=mv 2 /2d = (.6-kg)(4m/s) 2 /2(.005-m) F=960-N 960N.225lbs/1N=216lbs.

14. 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? v 1 =25-km/hr=7-m/s, v 2 = 100-km/hr=27.8-m/s K.E. 1 =1/2mv 2 =(.5)(1200kg)(7m/s) 2 =28,935J K.E. 2 =1/2mv 2 =(.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.

15. 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