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Work = The Amount of energy transferred by a force The word “Work” was first coined by French mathematician Gaspard-Gustave Coriolis in 1830 The work-Energy.

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Presentation on theme: "Work = The Amount of energy transferred by a force The word “Work” was first coined by French mathematician Gaspard-Gustave Coriolis in 1830 The work-Energy."— Presentation transcript:

1 Work = The Amount of energy transferred by a force The word “Work” was first coined by French mathematician Gaspard-Gustave Coriolis in 1830 The work-Energy theorem is: If an external force is applied to an object causing its kinetic energy to change, then work has been done. & W = F x d (Force)(Distance) Work = ∆E k (Change in kinetic energy) Engine, work & power

2 Horsepower: Coined by James Watt in ~1787 When a horse could turn a mill wheel 144 times an hour (2.4 times per minute) the wheel had a 12ft radius, so traveled 2.4x2xpix12ft in one minute. He figured intellectually that a horse could pull 180 pounds continuously, so: In reality a very optimistic value but it helped him sell steam engine. Power (HP)= Work / Time = (Force x Distance) / Time = (180lbs)x(2.4x2xpix12ft)/ Time = 32572 (ft.lbs)/min = 542.8 (ft.lbs)/sec Engine, work & power

3 Point of maximum volumetric efficiency Horsepower curve RPM0 100 500 400 300 200 0 10002000 3000 5000 400070006000 8000 HP peak Engine, work & power

4 Area under the curve = total work done. To maximize acceleration (and efficiency) run the engine as much as possible in this range. Engine, work & power Horsepower curve RPM0 100 500 400 300 200 0 10002000 3000 5000 400070006000 8000 HP peak

5 Spring Computation M M K Spring K Tire

6 Spring Computation KBKB M KAKA Total stiffness when 2 springs are in series:

7 Spring Computation KBKB M KAKA Total stiffness when 2 springs are in series: K A = 1600 lbs/in K B = 1200 lbs/in K A+B = (1600 x 1200)/(1600 + 1200) K A+B = 686 lbs/in

8 Spring Computation KCKC M KAKA Total stiffness when 2 springs are in series and in parallel: KBKB

9 Spring Computation Total stiffness when 2 springs are in series and in parallel: KCKC M KAKA KBKB K A = 1600 lbs/in K B = 700 lbs/in K C = 1200 lbs/in K A+B+C = ((1600+700) x 1200)/((1600+700) + 1200) K A+B+C = 789 lbs/in

10 Spring Computation Area under the curve = total work done. Spring Force - deflection Curve Distance 00.20.4 0.6 1.0 0.81.41.2 1.6 0 500 2500 2000 1500 1000 Force

11 Time -100 -75 -50 -25 0 25 50 75 100 051015202530 Amplitude Time Spring decaying oscillations Spring Under DampedSpring Normally DampedSpring Over Damped

12 Spring decaying oscillations “Spring Under Damped” Work = “Spring Normally Damped” Work = “Spring Over Damped” Work =

13 Spring decaying oscillations If Damping , then Spring Work 

14 Spring decaying oscillations Spring with no dampingSpring dampedShock damping work


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