2. Physics 151: Lecture 13, Pg 1 Physics 151: Lecture 12 l Today’s Topics: çReview Springs çReview Work - Energy Theorem. çProblems we can do with that.

3. Physics 151: Lecture 13, Pg 2 Definition of Work: F r Work (W) of a constant force F acting through a displacement  r is: F r rr W = F.  r = F  r cos  = F r  r See text: 7-1 Definition of Kinetic Energy : The kinetic energy of an object of mass (m) moving at speed (v) is: 1/2 m v 2 K = 1/2 m v 2 Work Kinetic-Energy Theorem:

4. Physics 151: Lecture 13, Pg 3 Question Planets go around the sun in elliptical orbits. The highly exaggerated diagram below shows a portion of such an orbit and the force on the planet at one position along that orbit. The planet is moving to the right. F T and F R are the components of the force parallel (tangential) and perpendicular (radial) to the orbit. The work they do is W T and W R at the position shown a. W T slows the planet down; W R speeds it up. b. W T slows the planet down; W R does no work on it. c. W T speeds the planet up; W R does no work on it. d. W T speeds the planet up; W R slows it down. e. W T does no work on it; W R speeds the planet up.

5. Physics 151: Lecture 13, Pg 4 Act 1 When a ball rises vertically to a height h and returns to its original point of projection, the work done by the gravitational force is : (a) 0 (b) -mgh (c) +mgh (d) -2mgh (e) +2mgh When a crate of mass m is dragged a distance d along a surface with coefficient of kinetic friction, then dragged back along the same path to its original position, the work done by friction is : (a) 0 (b) -u k mgd (c) +u k mgd (d) -2u k mgd (e) +2u k mgd Act 2

6. Physics 151: Lecture 13, Pg 5 Work Done Against Gravity l Consider lifting a box onto the tail gate of a truck. m h The work required for this task is, W = F. d =(mg)(h) (1) W = mgh

7. Physics 151: Lecture 13, Pg 6 Work Done Against Gravity l Now use a ramp to help you with the task. Is less work needed to get the box into the truck? (It’s “easier” to lift the box) m h

8. Physics 151: Lecture 13, Pg 7 Work Done Against Gravity m mgmg  mgsin  mgcos  F To push the box with constant speed, F = mgsin  The length of the ramp is h/sin  So the work done is, W = Fd = (mgsin  )(h/sin  ) W = mgh Same as before ! N h

9. Physics 151: Lecture 13, Pg 8 Lecture 13 - ACT 3 Work of Springs l I have a spring with k = 20 N/m. Its compressed 30 cm by a 200 gram mass. How much work is done by the spring in this process ? A) 0.6 JB) 6 JC) 0.9 JD) –0.9 J

10. Physics 151: Lecture 13, Pg 9 Act 4 Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their speeds at the instant of launch is: a)v A = 3/2 v B b) v A = (3/2 ) 1/2 v B c) v A = v B d) v B = (3/2 ) 1/2 v A e) v B = 3/2 v A

11. Physics 151: Lecture 13, Pg 10 At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety device to stop the train so that it will not go plowing through the station if the engineer misjudges the stopping distance. While waiting, you wonder what would be the fastest train that the spring could stop by being fully compressed, 3.0 ft (~ 1m). To keep the passengers as safe as possible when the spring stops the train, you assume that the maximum stopping acceleration of the train, caused by the spring, is g/2. You make a guess that a train might have a mass of 0.5 million kilograms. For the purpose of getting your answer, you assume that all frictional forces are negligible. You find the maximum speed the train can have is: (A) 10 mph ( Note: 1 m/s ~ 2 mph ) Example Problem

12. Physics 151: Lecture 13, Pg 11 Lecture 13, ACT 5 Work & Energy Two blocks having mass m 1 and m 2 where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. Which one will go farther before stopping ? (a) (b) (c) (a) m 1 (b) m 2 (c) they will go the same distance m1m1 m2m2

13. Physics 151: Lecture 13, Pg 12 Lecture 13, ACT 6 Work & Energy l You like to drive home fast, slam on your brakes at the bottom of the driveway, and screech to a stop laying rubber all the way. It’s particularly fun when your mother is in the car with you. You practice this trick driving at 20 mph and with some groceries in your car with the same mass as your mama. You find that you only travel half way up the driveway. Thus when your mom joins you in the car, you try it driving twice as fast. How far will you go this time ? (a)The same distance. Not so exciting. (b)  2 times as far (only 7/10 of the way up the driveway) (c) twice as far, right to the door. Whoopee! (d) four times as far. Crashes into house. Sorry Ma.

14. Physics 151: Lecture 13, Pg 13 Another Example A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown. The speed of the block increases from 4.0 m/s to 6.0 m/s in a displacement of 5.0 m. What work was done by the friction force during this displacement? a.–34 J b. –64 J c. –30 J d. –94 J e. +64 J

15. Physics 151: Lecture 13, Pg 14 Another Example 1.You have been hired to design a spring based toy pistol. The design criterion states that you have to be able to shoot an extremely aerodynamic bullet of mass 50 grams a distance of 20 meters. Unfortunately, there are not that many choices you can make, because the company insists on you using the left-over plastic gun-bodies from a previous, failed product. You do however have to design an appropriate spring. The gun body has a barrel of length 10 cm on the outside and 8 cm on the inside. The entire plastic body has a mass of 150g. You figure that the spring can fill the entire barrel, and can probably have a maximum compression to about half of its original length. What is the design of your spring ? (k=? ) k = 6.2 kN/m

16. Physics 151: Lecture 13, Pg 15 Recap of today’s lecture l Work/Energy Theorem  W =  K l Kinetic Energy çK = 1/2 mv 2