1. PHYSICS 111SI- WORK

2. Formulas  Work in the x direction = Force in the x direction * displacement in the x direction  W=F//S  Work is in joule (N * m)  W= delta KE =.5mVf^2 -.5mvi^2

3. Conceptual Check  In a baseball game, the catcher stops 90 mph pitch. What can be say about the work done by the cater on the ball?  Cather has done positive work  Cather has done negative work  Cather has done zero work

4. Answer:  Cater has done negative work  His stopping force is 180 degrees from the displacement of the ball Thus it is a negative force and does negative work

5. Concept 2  A ball tied to a string is being whirled around in a circle. What can you say about the work done by tension?  Tension does no work  Tension does negative work  Tension does positive work

6. Answer:  Tension does no work at all  If you draw picture, it is perpendicular to the movement of the ball

7. Last one:  You lift a book with your hand in such a way that it moves up at a constant speed and has no acceleration. While it is moving up, what is the total work done on the book?  Mg * delta x  F hand * delta x  (F hand + mg) * delta x  Zero  None of the above

8. Solution:  D. zero  The total net force is zero, so w=f//s equals zero

9. PRACTICE Problem  A child is pulling their sled across a flat surface. The child pulls forward with a force of 23 N at an angle of 30 degrees. The child need to take the sled a hundred meters. How much work does the child need to do this?

10. Solution:  W = F//S  F// = Force in the x direction  23 Cos 30 = Fx = F// = 19.91 N  Know the displacement is 100 m  W= 19.91(100) = 1991.85 joules or 1.99 kj

11. Problem:  Find the NET WORK done when a 10 kg box is pulled across a floor with a magnitude of 43 N at an angle of 15 degrees. The box travels 40 meters and the coefficient of kinetic friction is.12.

12. Solution:  Two ways to do this problem  1. You can find the total net force and multiple that by displacement  2. Find the individual work for the force of friction and the force forward, add them together, and them multiple by the displacement Net Force = ( 43 Cos 15) + (.12 * (10 * -9.81) = 40.5 + -11.77 = 28.728 N in the x direction W=(28.7 N)(40 m) = 1148 joules

13. Kinetic Energy  A 900-kg compact car moving at 60 mi/hr has approximately 320000 Joules of kinetic energy. Estimate its new kinetic energy if it is moving at 30 mi/hr.

14. Answer:  You know that the mass =900 kg  Vi = 30 mi/ hr * 1 hr / 3600 sec * 1609.34 meter / mi  Vi= 13.41 meters per second  Use equation KE =.5 m v^2  KE f =.5 (900)(13.41)^2 =80,923 Joules

15. Challenge Problem  Jerome jumps up and he raises his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.  What is the work done by Jerome?  What is the power generated by him?

16. Solution:  Work = F//S  F=ma = 102 kg * a Acceleration = v.t  S=2.29 meters  Work = 2.29 meters ( 102 kg * 9.81) = 2291 Joules  Power = work / time P = 2291 j / 1.32 sec = 1735 wyatts or j / sec