Don’t forget you can do this as area under the curve!!! ½ base * height ½ x * F ½ x (kx) ½ kx 2
a. k = 64 N/m b. W = 200 J c. F = 122 N d. P = 15.4 watts
Transcript
1. Work & EnergyIn the past… v, a, x, t  How things move, Kinematics F, a, m  What makes them move, DynamicsNow we will look at WHY they move!!! Energy!Energy  The ability to do work.
2. Work = Force x Displacement1 Joule = 1 Newton x 1 Meter (jewel)1. Object must move.2. Force & Displacement must be on the same plane3. Don’t forget air friction is negligible.
3. Anything that puts a force on an object displacing it will cause work. Which of the following do work on the box? gravity  No. Doesn’t move up or down normal force  No. Doesn’t move up or down you pulling it  Yes, but only the x component friction  YesPositive  energy is + if it is going into the object/systemNegative  energy is – if it is coming out of the object/system
4. How much work will the road do on an 1800 kg car when its brakes are applied, if the coefficient of friction between the road and the wheels is 0.5 and the car skids 6 m? W = F d = Ff d = µ Fg d = µ m g d = (0.5) (1800) (9.8) (6) W = -52, 920 J
5. F vs d Constant Forces How do you find work onF F vs d graph? Work = F d = Area under the curve d F vs d Nonconstant forces ΔW = F Δd F W = F ∫d d
6. Gravitational Potential Kinetic EnergyEnergy  energy due to motion  energy due to position KE = ½ mv 2PE = mgh Ex. How much KE does an 1800Ex. You lift a 1.2 kg book from kg car going 25 mph (11.2 m/s)the first floor to your social have?studies class on the 2nd floor 5m up. How much potential KE = ½ mv2energy does the book have? = ½(1800)(11.2)2 = 113 000 JPE =mgh = (1.2) (9.8) (5) How much work would friction = 58.8 J need to do to stop it? W = KE = -113 000 JHow much work did you do?Conservation!!! W = PE = 58.8 J
7. Law of All the energy in = All the energy outConservatio W + PE + KE = PE + KE (+W)n of Energy (Work out is done by friction. If no Energy friction, then no work out.)cannot becreated or A 600 kg roller coaster car is lifted todestroyed the top of the first hill, 55 m above the ground.a. How much potential energy does it have? PE = m g hb. How much work was done to get the cart to the top of the first hill? W = PE PE = KE = ½ mv2c. How fast is it going at the bottom of the first hill?d. If the second hill is 40 m high, then how fast will the cart be going when it crests the hill? PE = PE + KE
8. Sample ProblemA disgruntled physics student dropsher book off a 4 story building (12 m),how fast is the book going before ithits the ground?h = 12 mm = 1.7 kg h=Energy in = Energy out 12 mPE + KE = PE + KEDouble check with kinematics!
9. Describe the energy transfer in the following Different Scenarios• Dropping an object off a building• Throwing an object off a building• Car being slowed down by friction• You throwing a ball• A bullet shot; then embedded in a tree• You lifting your backpack up to math
10. Work Energy Theorem W = ΔKE W = ΔPEIn order to change any type of energy, work must be done.
11. Power = Work/TimeTells you how much energy you use in a certain amount of time.Metric Unit: WattsEnglish Unit: Horsepower.746 Watts = 1 HP
12. A school bus pulls into an intersection. A cartraveling 35 km/h approaches and hits a patchof ice. The driver locks the brakes causing thecar to slide toward the intersection. If the caris originally 26 m away and the coefficient offriction between the car’s tires and the icy roadis 0.25, does the car hit the bus and poorinnocent school children lose their lives … ordoes the car stop just in the nick of time lettingthe little children grow up to do physicsproblems involving school buses and icyroads?
13. Variable Forces Force the spring applies is directly proportional toSprings how far it is stretched. Fαx F=kx k is called the spring constant FS F if k is large, spring is stiff if k is small, spring is loose FS F Hooke’s Law (Robert Hooke) Xi Xf
14. Energy in a Spring F + Fi F kx 1How would you find the F= = = = kx 2 2 2 2 work you put into stretching a spring? W = Fd = ( 1 kx ) xW=Fd 2But the Force changes over W = 1 kx 2 2 the distance…So let’s find the average force… When you do work on a spring, where does that energy go? Potential Energy!!! W = PEs = ½kx2
15. Sample ProblemA woman weighing 600 N steps on abathroom scale containing a spring. Thespring is compressed 1.0 cm under herweight. Find the force constant of thespring and the total work done on it duringcompression. F = kx k = F/x = 600/0.01 = 60,000 N/m W=½ kx2=½ (60,000) (0.01)2 W = 3.0 Nm = 3.0 J
16. Garage DoorA large garage door spring is stretched a distance of 2.50 m when a force of 160 N is applied. Find:a. the spring constantb. the work done on the springc. the force needed to stretch the spring 1.90 md. the power used if the spring is stretched 1.20 m in 3.00 secs.
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