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UNIT 3 Forces and the Laws of Motion Monday October 24th FORCES & THE LAWS OF MOTION 2 TODAY’S AGENDA Monday, October 24 Laws of Motion Mini-Lesson: Everyday Forces (2nd Law Problems) UPCOMING… Thurs: Newton’s 2nd Law Lab Fri: Quiz #2 2nd Law Problem Mon: Test Review Tue: TEST #4 Fy 0 N m1g Force Lab Notes Fx m1a T m1a Forces on m1 N F ma T T m 2 g m 2a m1 T m1g m2 m2g Forces on m2 m 2 g T m 2a m1a = T = m2g – m2a m1a m 2g m 2a a m 2g m1 m 2 23 Everyday Forces A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. a) Find the μk between the box and the ramp. b) What acceleration would a 175 kg box have on this ramp? 25 A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. a) Find the μ between the box and the ramp. ΣFy = 0 FN = mgcos(25°) = 667 N ΣFx ≠ 0 FNET = ma = mgsin(25°) - Ff FN FNET = 270 N = 311- Ff Ff Ff = µFN = µ(667N) = 41N mgcos(25°) mgsin(25°) µ = .0614 mg 26 A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. ΣFx ≠ 0 b) What acceleration would a 175 kg box have on this ramp? Ff = µFN FN FNET = ma ma = mgsin(25°) - Ff Ff ma = mgsin(25°) – μmgcos(25˚) mgcos(25°) mg mass does not matter, the acceleration is the same!! mgsin(25°) 27 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? FN vi = 0 F t=? 4.00 m Ffk Fg 28 FN 90.0N 77.9 N Ffk = 30˚ 45.0 N 735.8 N 1. Draw a free-body diagram to find the net force. 2. Convert all force vectors into x- and y- components. 29 FN = 781 N 90.0N 77.9 N Ffk = 30˚ 45.0 N 735.8 N 3. Is this an equilibrium or net force type of problem? Net force ! 4. The sum of all forces in the y-axis equals zero. 5. Solve for the normal force. FN = 45.0 + 735.8 N FN = 781 N 30 FN = 781 N 90.0N 77.9 N Ffk = 30˚ 45.0 N 44.5 N 735.8 N 6. Given the μk = 0.057, find the frictional force. μkFN = Ff (0.057) 781 N = 44.5 N Ff = 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/s2 31 FN = 781 N 90.0N 77.9 N Ffk = 30˚ 45.0 N a = .445 m/s2 44.5 N 735.8 N 8. Which constant acceleration equation has a, vi, x, and t? t = 4.24 s 32 END 13