01-25-08 - Intro To Power
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The document discusses power and how it relates to work. It defines power as the rate at which work is done, and provides the formula that power is equal to work divided by time. It gives examples of calculating power from examples of objects being moved over distances in certain times, applying force. Units of power like watts and horsepower are also explained.
01-25-08 - Intro To Power
- 1. Power! We’ve talked about “work”, now let’s talk about “power”
- 2. “Work” Refresher Remember that the formula for finding the amount of work done upon an object is: W = (F)(d)(cos Θ ) W = the work done F = the force required to cause displacement d = displacement of the object Θ = the angle between the direction of the force and the direction of the displacement
- 3. Definition of Power Common definition usually includes something about strength Physics definition - the timed rate at which work is done depends directly upon the work and inversely upon the time to do that work If you increase work, then you increase power BUT If you increase time, then you decrease power
- 4. How “Hard” Are You Working? The rate at which work is done is called power : (Power) = (Work Done) / (Time Spent Working) P = W / t Power is “how hard” something is working.
- 5. Units for power : 1 J 1 s = 1 W one watt (W) equals one Joule (J) per second (s) Joules per seconds = Watts
- 6. Kilowatts and Horsepower Another common unit of power (not used in science, but used in everyday life) is the “horsepower” – basically the rate at which a (very powerful, very healthy) horse can do work over a 10 hour “work day.” 1 horsepower (hp) = 746 W or 0.746 kW We say 1,000 watts (W) = 1 kilowatt (kW), the same way that 1,000 meters = 1 kilometer
- 7. How fast can your car do work? A compact car may have a 120 hp engine. That means the car’s power is (120 hp)(746 W/hp) = 89,520 W So a typical compact car can do 89,520 J worth of work each second.
- 8. Remember this Example of Work? You are pushing a very heavy stone block (200 kg) across the floor. You are exerting 620 N of force on the stone, and push it a total distance of 20 m before you get tired and stop. How much work did you just do? W = (620 N)(20 m)(cos 0 °) = 12,400 J
- 9. Power Example Let’s say that it took us 40 s to move that 200 kg stone block the 20 m. Remember that we did 12,400 J of work on the stone block Given Information: W = 12,400 J t = 40 s Basic Equation: P = W/t Working Equation: P = 12,400 J / 40 s P = 310 J/s Since it took us 40 s to move the block, we were doing 310 J of work per second OR generating 310 W of power
- 10. Remember the Weightlifter? A weightlifter lifts a barbell with a mass of 280 kg a total of 2 meters off the floor. What is the minimum amount of work the weightlifter did? The barbell is “pulled” down by gravity with a force of (280 kg)(10 m/s 2 ) = 2,800 N So the weightlifter must exert at least 2,800 N of force to lift the barbell at all. If that minimum force is used, the work done will be: W = (2,800 N)(2 m) = 5,600 J
- 11. Power What about the weightlifter? How much power does the weightlifter generate if it takes him 2 seconds to lift the weight? (W = 5,600 J) P = 5,600 J / 2 s = 2,800 J/s = 2,800 W We say 1,000 watts = 1 kilowatt So the weightlifter’s power was 2.8 kW
- 12. Fast work isn’t more work Go back to our 200 kg block example. Remember that when it took us 40 s to push the block the 20 m when we applied a force of 620 N, that implied that we had a power output of 310 W. If we exerted the same force (620 N) and pushed the block the same distance (20 m), but took half as long to do so (20 s), our power output would double to 620 W.
- 13. Fast work isn’t more work *But* Notice that the total work done doesn’t change – we still exerted 620 N of force over a distance of 20 m. So increasing power output doesn’t mean you’re doing more work, it means you’re doing the work faster .
- 14. Game Time!!! Break into 4 groups (with even amounts of people in each group) A question will be projected onto the board and 2 groups will race to complete the question correctly The winning group will get a reward In order to receive the reward, groups must include all steps needed to solve a word problem
- 15. Ready??? Groups 1 and 3 will start with question 1, then Groups 2 and 4 will work question 2, etc Groups not working on questions, must stay quiet Rewards will go to those actively participating with their group (writing on the board, working it out on separate paper, paying attention, etc)
- 16. Question 1 – Groups 1 and 3 A box that weighs 575 N is lifted a distance of 20.0 m straight up. The job is done in 10.0 s. How much power is generated? Include a diagram, given information, B.E., W.E., and answer Answer: 1150 W or 1.15 kW
- 17. Question 2 – Groups 2 and 4 An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35 s. How much force (F L ) does the motor exert? Include a diagram, given information, B.E., W.E., and answer Answer: 130,000 N
- 18. Question 3 – Groups 1 and 3 Jose lifts a 20 kg mass to a height of 2.0 m in 5.0 s. Sue lifts 30 kg to a height of 1.5 m in 8.0 s. Who did more work? Who had more power? Include a diagram, given information, B.E., W.E., and answer Answer: Sue did more work (450 J), less power (56 W) Jose generated more power (80 W), less work (400 J)
- 19. Question 4 – Groups 2 and 4 A 185 kg refrigerator is loaded into a moving van by pushing it up a 10.0 m ramp at an angle of inclination of 11.0 °in 20 s. How much work is done by the pusher? How much power was generated? Include a diagram, given information, B.E., W.E., and answer Answer: W = 3460 J P = 173 W
- 20. And the WINNER is… Everyone!! Because everyone helped to work the problems you have all been preparing for the test Good job!