1. UNIT 3 Forces and the Laws of Motion

2. Monday October 24 th 2 FORCES & THE LAWS OF MOTION

3. TODAY’S AGENDA  Laws of Motion  Mini-Lesson: Everyday Forces (2 nd Law Problems) UPCOMING…  Thurs:Newton’s 2 nd Law Lab  Fri:Quiz #2 2 nd Law Problem  Mon:Test Review  Tue:TEST #4 Monday, October 24

4. m1m1 m2m2 N m1gm1g T T m2gm2g Forces on m 1 Forces on m 2 23 m 1 a = T = m 2 g – m 2 a Force Lab Notes

5. Everyday Forces a) Find the μ k between the box and the ramp. b)What acceleration would a 175 kg box have on this ramp? A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s 2. 25

6. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s 2. a) Find the μ between the box and the ramp. FNFN mg mgcos(25°) FfFf mgsin(25°) ΣF y = 0 F N = mgcos(25°) = 667 N F NET = ma = mgsin(25°) - F f F NET = 270 N = 311- F f F f = µF N = µ(667N) = 41N µ =.0614 ΣF x ≠ 0   26

7. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s 2. b) What acceleration would a 175 kg box have on this ramp? FNFN mg mgcos(25°) FfFf mgsin(25°) F NET = ma F f = µF N ΣF x ≠ 0 ma = mgsin(25°) - F f ma = mgsin(25°) – μmgcos(25˚) mass does not matter, the acceleration is the same!! 27

8. Everyday Forces A 75.0-kg box is pushed with a 90.0N exerted downward at a 30˚ below the horizontal. If the coefficient of kinetic friction between box and the floor is 0.057, how long does it take to move the box 4.00m, starting from rest? A 75.0-kg box is pushed with a 90.0N exerted downward at a 30˚ below the horizontal. If the coefficient of kinetic friction between box and the floor is 0.057, how long does it take to move the box 4.00m, starting from rest? F  4.00 m FNFN FgFg F fk t = ? v i = 0 28

9. 90.0N  FNFN 735.8 N F fk  = 30˚ 1. Draw a free-body diagram to find the net force. 2. Convert all force vectors into x- and y- components. 77.9 N 45.0 N 29

10. 90.0N  FNFN 735.8 N F fk  = 30˚ 3. Is this an equilibrium or net force type of problem? 77.9 N 45.0 N 4. The sum of all forces in the y-axis equals zero. 5. Solve for the normal force. Net force ! F N = 45.0 + 735.8 N F N = 781 N = 781 N 30

11. 90.0N  FNFN 735.8 N F fk  = 30˚ 6. Given the μ k = 0.057, find the frictional force. 77.9 N 45.0 N μ k F N = F f = 781 N (0.057) 781 N = 44.5 NF f = 44.5 N 44.5 N 7. Given this is a net force problem, net force equals m times a. 77.9 N – 44.5 N = (75 kg) a a =.445 m/s 2 31

12. 90.0N  FNFN 735.8 N F fk  = 30˚ 8. Which constant acceleration equation has a, v i, x, and t? 77.9 N 45.0 N = 781 N 44.5 N t = 4.24 s a =.445 m/s 2 32

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