1. Work and Energy 1.Work Energy  Work done by a constant force (scalar product)  Work done by a varying force (scalar product & integrals) 2.Kinetic Energy Chapter 6: Work and Energy Work-Energy Theorem

2. Work and Energy Work by a Baseball Pitcher A baseball pitcher is doing work on the ball as he exerts the force over a displacement. v 1 = 0 v 2 = 44 m/s

3. Work and Energy Work Done by a Constant Force (I) Work (W)  How effective is the force in moving a body ? W [Joule] = ( F cos  ) d  Both magnitude (F) and directions (  ) must be taken into account.

4. Work and Energy Work Done by a Constant Force (II) Example: Work done on the bag by the person..  Special case: W = 0 J a) W P = F P d cos ( 90 o ) b) W g = m g d cos ( 90 o )  Nothing to do with the motion

5. Work and Energy Example 1A A 50.0-kg crate is pulled 40.0 m by a constant force exerted (F P = 100 N and  = 37.0 o ) by a person. A friction force F f = 50.0 N is exerted to the crate. Determine the work done by each force acting on the crate.

6. Work and Energy Example 1A (cont’d) W P = F P d cos ( 37 o ) W f = F f d cos ( 180 o ) W g = m g d cos ( 90 o ) W N = F N d cos ( 90 o ) 180 o 90 o d F.B.D.

7. Work and Energy Example 1A (cont’d) W P = 3195 [J] W f = - 2000 [J] (< 0) W g = 0 [J] W N = 0 [J] 180 o

8. Work and Energy Example 1A (cont’d) W net =  W i = 1195 [J] (> 0)  The body’s speed increases.

9. Work and Energy Example 2 A car traveling 60.0 km/h to can brake to a stop within a distance of 20.0 m. If the car is going twice as fast, 120 km/h, what is its stopping distance ? (a) (b)

10. Work and Energy Example 2 (cont’d) (1) W net = F d (a) cos 180 o = - F d (a) = 0 – m v (a) 2 / 2  - F x (20.0 m) = - m (16.7 m/s) 2 / 2 (2) W net = F d (b) cos 180 o = - F d (b) = 0 – m v (b) 2 / 2  - F x (? m) = - m (33.3 m/s) 2 / 2 (3) F & m are common. Thus, ? = 80.0 m

11. Work and Energy Work-Energy Theorem W net = F net d = ( m a ) d = m [ ( v 2 2 – v 1 2 ) / 2d ] d = (1/2) m v 2 2 – (1/2) m v 1 2 = K 2 – K 1

12. Work and Energy Spring Force (Hooke’s Law) F S (x) = - k x FPFP FSFS Natural Length x > 0 x < 0 Spring Force (Restoring Force): The spring exerts its force in the direction opposite the displacement.

13. Work and Energy Work Done to Stretch a Spring x 2 W = F P (x) dx x 1 F S (x) = - k x W Natural Length FPFP FSFS

14. Work and Energy l b W = F || dl l a Work Done by a Varying Force  l  0

15. Work and Energy Example 1A A person pulls on the spring, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the person do ? If, instead, the person compresses the spring 3.0 cm, how much work does the person do ?

16. Work and Energy (a) Find the spring constant k k = F max / x max = (75 N) / (0.030 m) = 2.5 x 10 3 N/m (b) Then, the work done by the person is W P = (1/2) k x max 2 = 1.1 J (c) x 2 = 0.030 m W P = F P (x) d x = 1.1 J x 1 = 0 Example 1A (cont’d)

17. Work and Energy Example 1B A person pulls on the spring, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the spring do ? If, instead, the person compresses the spring 3.0 cm, how much work does the spring do ?

18. Work and Energy (a) Find the spring constant k k = F max / x max = (75 N) / (0.030 m) = 2.5 x 10 3 N/m (b) Then, the work done by the spring is (c) x 2 = -0.030 m  W S = -1.1 J x 2 = -0.030 m W S = F S (x) d x = -1.1 J x 1 = 0 Example 1B (cont’d)

19. Work and Energy Example 2 A 1.50-kg block is pushed against a spring (k = 250 N/m), compressing it 0.200 m, and released. What will be the speed of the block when it separates from the spring at x = 0? Assume  k = 0.300. (i) F.B.D. first ! (ii) x < 0 F S = - k x

20. Work and Energy (a) The work done by the spring is (b) W f = -  k  F N (x 2 – x 1 ) = -4.41 (0 + 0.200) (c) W net = W S + W f = 5.00 - 4.41 x 0.200 (d) Work-Energy Theorem: W net = K 2 – K 1  4.12 = (1/2) m v 2 – 0  v = 2.34 m/s x 2 = 0 m W S = F S (x) d x = +5.00 J x 1 = -0.200 m Example 2 (cont’d)