Potential energy. Web page:

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1 Potential energy Announcements: CAPA homework due at 10pm today New CAPA assignment available at 5pm. Grading questions on Midterm connected with how scantron sheets filled out will need to see Professor Dan Dessau. Starting Chapter 8 in H+R Web page:

2 Review of Chapter 6 Work is Kinetic energy is Work-energy theorem is Instantaneous power is

3 When: Monday, March 5, 2012, at 6 p.m. Where: UMC 235 (hall right of Reception Desk) RSVP: By March 1 st to laprogram@colorado.edu Refreshments will be served, while they last. Come to the LA Info Session to learn more about becoming a Learning Assistant. Applications for Fall 2012 available March 5 19 Goto: Get more information from faculty and LAs in these departments: Applied Math Math MCDBiology EBIO Education Chemistry Physics Astronomy ATOC Mech Engineering

4 Learning Assistants (LAs) Get Paid - $10/hr x 10hr/wk x15 weeks = $1500 Get to choose which courses you will be in Physics 1110, Sound and Music, Physics of Everyday Life Physics 1010 and opportunities in other departments. It's meant for people who are thinking about the possibility of teaching at some point in their life. It's not restricted for future teachers.

5 Clicker question 1 Two marbles, one twice as heavy as the other, are dropped to the ground from the top of a building. (Assume no air resistance.) Just before hitting the ground, the heavier marble has A. the same kinetic energy as the lighter marble B. twice the kinetic energy of the lighter marble C. half the kinetic energy of the lighter marble D. four times the kinetic energy of the lighter marble E. one-quarter the kinetic energy of the lighter marble

6 Clicker question 1 Two marbles, one twice as heavy as the other, are dropped to the ground from the top of a building. (Assume no air resistance.) Just before hitting the ground, the heavier marble has A. the same kinetic energy as the lighter marble B. twice the kinetic energy of the lighter marble C. half the kinetic energy of the lighter marble D. four times the kinetic energy of the lighter marble E. one-quarter the kinetic energy of the lighter marble Objects in free fall (with no air resistance) have acceleration equal to so both marbles hit the ground with the same speed. Since kinetic energy is, the marble that is twice as heavy will have twice the kinetic energy

7 Example 2 (with Power) A kg car requires 12 hp to cruise at a steady 80km/h on a level road. What would be the power required to move up a 10 degree incline at the same speed? Assume the total frictional force due to the road and air resistance is fixed. At constant speed the wheels (actually the road) provide a forward force F that just balances the rolling friction f r of the tires and the drag force f d due to the air. This means that f T =f r + f d. Hence the power delivered to the wheels of the car is P = f T v. Converting to SI units gives v=80 km/h=(80 x 10 3 m)/(3600 s) = 22.2 m/s and 12 hp = 12 x 746 =8.95 x 10 3 W. Thus, f T = P v = W 22.2m /s = 403N Now ready for the incline..

8 Example 2 cont. f T = P v = W 22.2m /s = 403N To go up an incline at constant speed, requires the car to have a force F = f T + mgsin10. The power required is P = F v = (403N kg 9.8m /s 2 sin10 )(22.2m /s) P = W = 62.7hp Note this power is 5x higher than on a level road!

9 Clicker question 2 A sports car accelerates from zero to 30 mph in 2 seconds. How long does it take to accelerate from zero to 60 mph, assuming that the power output of the engine is independent of velocity and neglecting friction and air resistance? A. 4 seconds B. 6 seconds C. 8 seconds D. 16 seconds E. None of the above

10 Clicker question 2 A sports car accelerates from zero to 30 mph in 2 seconds. How long does it take to accelerate from zero to 60 mph, assuming that the power output of the engine is independent of velocity and neglecting friction and air resistance? A. 4 seconds B. 6 seconds C. 8 seconds D. 16 seconds E. None of the above Doubling the speed quadruples the kinetic energy Therefore, 4 times more work is needed to accelerate from 0 to 60 mph than 0 to 30 mph. Since the power is constant it will take 4 times longer to apply 4 times more work P = dw dt

11 Clicker question 3 A 1 kg mass is moved part way around a square loop as shown. The square is 1 m on a side and the final position of the mass is 0.5 m below its original position. Assume g = 10 m/s 2. What is the work done by the force of gravity during this journey? A. 10 J B. 5 J C. 0 J D. 5 J E. 10 J 1.0 m up 0.5 m start finish

12 Clicker question 3 A 1 kg mass is moved part way around a square loop as shown. The square is 1 m on a side and the final position of the mass is 0.5 m below its original position. Assume g = 10 m/s 2. What is the work done by the force of gravity during this journey? A. 10 J B. 5 J C. 0 J D. 5 J E. 10 J 1.0 m up 0.5 m start finish No work from the horizontal segments since force and displacement are perpendicular Upward segment: Downward segment:

13 Clicker question 3 A 1 kg mass is moved part way around a square loop as shown. The square is 1 m on a side and the final position of the mass is 0.5 m below its original position. Assume g = 10 m/s 2. What is the work done by the force of gravity during this journey? A. 10 J B. 5 J C. 0 J D. 5 J E. 10 J 1.0 m up 0.5 m start finish Can also think of the 0.5 m upward segment being canceled out by half of the 1 m downward segment leaving just a 0.5 m downward segment Total work:

14 Clicker question 4 A 1 kg mass is moved on a totally crazy path that ends up in the same place as in the previous problem. Is the work done by gravity during this journey the same or different compared to the previous path? A. Same B. Different C. Impossible to tell 1.0 m up start finish 0.5 m

15 Clicker question 4 A 1 kg mass is moved on a totally crazy path that ends up in the same place as in the previous problem. Is 1.0 m up the work done by gravity during this start journey the same or different 0.5 m compared to the previous path? A. Same finish B. Different C. Impossible to tell No work comes from any horizontal motion With the exception of the 0.5 m of difference between the start and end, all other upward and downward motion cancels out Therefore, the work done by gravity is the same!

16 Gravity is a conservative force The reason the work done by gravity is independent of the path taken is because gravity is a conservative force. Some properties of conservative forces Work done by a conservative force is independent of path taken Work done by a conservative force over a closed loop (starting and ending at the same place) is 0 A conservative force only depends on position (or is constant)

17 Clicker question 5 A 1 kg mass is slid on a table along two different paths as shown which both end up in the same place. Is the magnitude of the work due to friction A. the same for both paths B. greater for path A C. greater for path B Path A start finish Path B start finish

18 Clicker question 5 A 1 kg mass is slid on a table along two different paths as shown which both end up in the same place. Is the magnitude of the work due to friction A. the same for both paths B. greater for path A C. greater for path B Path A start finish start The force of friction is the same in both cases, µ k N. The frictional force is always opposite the motion so work due to friction is µ k Nd where d is the total distance traveled which is greater for path B so friction does more work in path B. Friction is not a conservative force! Path B finish

19 2.5 m N mg Newton s 2 nd law: Back to sliding blocks A block starts at rest at a height of 2.5 m and slides down a frictionless inclined plane. What is the block velocity when it reaches the bottom? Solution 1 The block slides a distance from Therefore Does not depend on!

20 2.5 m N mg Yet another sliding block Need the net force in the direction of motion: A block starts at rest at a height of 2.5 m and slides down a frictionless inclined plane. What is the block velocity when it reaches the bottom? Solution 2: work-energy theorem The block slides a distance Total work is From work-energy theorem from Same as previous result!

21 Potential energy We have found that the work due to gravity is where y 1 is the initial height and y 2 is the final height. One more rule about conservative forces is that we can define a quantity called potential energy for them. We define gravitational potential energy as. With this definition: If gravity is the only force contributing to the work then and so by the work energy theorem we arrive at K 2 -K 1 = -(U 2 -U 1 ) U 2 + K 2 = U 1 +K 1 Conservation of Mechanical Energy