1. Power Chapter 8.2 1

2. What determines the amount of work done by an object? The force on the object times its displacement. Time does not affect the amount of work done. 2

3. Power Power is the rate at which work is done. P = W/t P = power = W (Watts) W = work = J (Joules) t = time=s 3

4. A car engine has 50 horsepower and can make a car go from 0 mph to 60 mph in 18 seconds. How fast can a 150 horsepower engine make a car go from 0 mph to 60 mph? 18 s/3 = 6 s 4

5. How can a person be more powerful than another person? A person can be more powerful by exerting more force or by performing work in a shorter period of time. 5

6. Bill and Ted are climbing a wall that is 20 feet high. Bill uses a rope to climb the wall and took a whole minute to get to the top. Ted uses a ladder and climbs the wall in 25 s. Assuming Bill and Ted weighs the same, who does more work climbing the wall? Explain. They both did the same amount of work because they pulled the same weight to the same vertical distance. 6

7. Bill and Ted are climbing a wall that is 20 feet high. Bill uses a rope to climb the wall and took a whole minute to get to the top. Ted uses a ladder and climbs the wall in 25 s. Who generated more power? Explain. Ted generated more power because he completed the same amount of work in a shorter period of time. 7

8. Jack and Jill went up a hill to fetch a pail of water. Jack is twice as massive as Jill but took twice as long to reach the top of the hill compared to Jill. Who did more work going up the hill? Explain. Jack did more work because he exerted more vertical force (Fg = weight) going up the same hill. 8

9. A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power. Show your calculations. P = W/t = 0.50 J / 2 s = 0.25 W 9

10. When doing a chin-up, a physics student lifts her 42.0- kg body a distance of 0.25 meters in 2 seconds. What is the power delivered by the student's biceps? Show your calculations. F = student’s weight = mg = (42.0 kg)(9.8 m/s 2 ) = 412 N W = Fd = (412 N)(0.25 m) = 103 J P = W/t = 103 J / 2 s = 52 W 10

11. When doing a chin-up, a physics student lifts her 42.0- kg body a distance of 0.25 meters in 2 seconds. How would the amount of power delivered by the student change if she lost 2.0 kg and performed the same chin-up? Show your calculations. F = student’s weight = mg = (40.0 kg)(9.8 m/s 2 ) = 392 N W = Fd = (392 N)(0.25 m) = 98 J P = W/t = (98 J) / 2.0 s = 49 W 11