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UNIT 3 Forces and the Laws of Motion

UNIT 3 Forces and the Laws of Motion. Mon day October 24 th. FORCES & THE LAWS OF MOTION. TODAY’S AGENDA. Mon day , October 24. 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

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UNIT 3 Forces and the Laws of Motion

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  1. UNIT 3Forces and the Laws of Motion

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

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

  4. N T m1 T m1g m2 m2g Force Lab Notes Forces on m1 m1a = T = m2g – m2a Forces on m2 23

  5. 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. What acceleration would a 175 kg box have on this ramp? 25

  6. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2. ΣFy = 0 a) Find the μ between the box and the ramp. 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°)  µ = .0614 mg mgsin(25°)  26

  7. 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°) mass does not matter, the acceleration is the same!! mg mgsin(25°) 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? FN t = ? vi = 0 F  4.00 m Ffk Fg 28

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

  10. = 781 N 90.0N  77.9 N 45.0 N FN 735.8 N  = 30˚ 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. Ffk FN = 45.0 + 735.8 N 5. Solve for the normal force. FN = 781 N 30

  11. = 781 N 90.0N  77.9 N 45.0 N 44.5 N FN 735.8 N  = 30˚ 6. Given the μk = 0.057, find the frictional force. μkFN = Ff (0.057) 781 N = 44.5 N Ff = 44.5 N Ffk 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

  12. = 781 N 90.0N  77.9 N a = .445 m/s2 45.0 N 44.5 N FN 735.8 N  = 30˚ 8. Which constant acceleration equation has a, vi, x, and t? t = 4.24 s Ffk 32

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