Net Force Example 1 Using Weight What is

Net Force – Example 1 Using Weight What is the acceleration? 35 N 5. 0 kg F = ma 35 N – 49. 05 N = (5. 0 kg)a -13. 95 N = (5. 0 kg)a a = -2. 79 m/s/s TOC

What is the force? Net Force – Example 2 F=? Using Weight F = ma F – 78. 48 N = (8. 0 kg)(+3. 5 m/s/s) F = 106. 48 N 8. 0 kg (Accelerating upwards at 3. 5 m/s/s) TOC

Whiteboards: Using Weight 1|2|3|4|5 TOC

Find the acceleration: F = ma, weight = (8. 0 kg)(9. 81 N/kg) = 78. 48 N down Making up + <100. N - 78. 48> = (8. 0 kg)a 21. 52 N = (8. 0 kg)a a = 2. 69 m/s/s +2. 69 m/s/s 100. N 8. 00 kg W

Find the acceleration: F = ma, wt = (15. 0 kg)(9. 81 N/kg) = 147. 15 N down <120. N – 147. 15 N> = (15. 0 kg)a -27. 15 N = (15. 0 kg)a a = -1. 81 m/s/s It accelerates down -1. 81 m/s/s 120. N 15. 0 kg W

Find the force: F = ma, wt = (16 kg)(9. 81 N/kg) = 156. 96 N down <F – 156. 96 N> = (16. 0 kg)(+1. 5 m/s/s) F – 156. 96 N = 24 N F = 180. 96 N ≈ 180 N F 16 kg a = 1. 5 m/s/s (upward) +180 N W

Find the force: F = ma, wt = 1177. 2 N downward <F – 1177. 2 N> = (120. kg)(-4. 50 m/s/s) F – 1177. 2 N = -540 N F = 637. 2 N ≈ 637 N F 120. kg a = -4. 50 m/s/s (DOWNWARD) +637 N W

This box is going downwards at 22. 0 m/s and is stopped in a distance of 1. 85 m. What must be the upwards force acting on it to stop it? First, suvat: s = -1. 85 m, u = -22. 0 m/s, v = 0, a = ? use v 2 = u 2 + 2 as, a = +130. 81 m/s/s F = ma, wt = 1177. 2 N downward <F – 1177. 2 N> = (120. kg)(+130. 81 m/s/s) F – 1177. 2 N = 15697 N F = 16874. 5 ≈ 16, 900 N +16, 900 N F 120. kg W

If this box is initially at rest, what is its displacement in 5. 00 seconds? 14. 0 N 2. 10 kg -39. 3 m W

A falling 52. 0 kg rock climber hits the end of the rope going 13. 5 m/s, and is stopped in a distance of 4. 20 m. What was the average force exerted to stop them? F = ma, a = +21. 696 m/s/s (from kinematics) wt = 510. 12 N downward <F – 510. 12 N> = (52. 0 kg)(+21. 696 m/s/s) F = 1638. 3 N +1640 N F 52. 0 kg W

A 65. 0 kg dumbwaiter is going up at 5. 80 m/s and is brought to rest in a distance of 1. 90 m What is the tension in the cable supporting it as it is stopping? F = ma, a = -8. 853 m/s/s (from kinematics) wt = 637. 65 downward <F – 637. 65 N> = (65. 0 kg)(-8. 853 m/s/s) F = 62. 23 N +62. 2 N F 65. 0 kg W

Find the force: F = ma, wt = 1176 N downward <F – 1177. 2 N> = (120. kg)(-4. 50 m/s/s) F – 1177. 2 N = -540 N F = 637. 2 N • Relationship between tension, weight and acceleration • Accelerating up = more than weight (demo, elevators) • Accelerating down = less than weight (demo, elevators, acceleration vs velocity) • Climbing ropes F 120. kg a = -4. 50 m/s/s (DOWNWARD) W

Find the mass: F = ma, wt = m(9. 81 m/s/s) downward <13. 6 – m(9. 81 m/s/s)> = m(-1. 12 m/s/s) 13. 6 N = m(9. 81 m/s/s) - m(1. 12 m/s/s) 13. 6 N = m(9. 81 m/s/s-1. 12 m/s/s) 13. 6 N = m(8. 69 m/s/s) m = 1. 565017261 kg ≈ 1. 57 kg 13. 6 N m a = 1. 12 m/s/s (downward) 1. 57 kg W