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Net Force – Example 1 Using Weight 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 5.0 kg 35 N What is the acceleration?

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Presentation on theme: "Net Force – Example 1 Using Weight 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 5.0 kg 35 N What is the acceleration?"— Presentation transcript:

1 Net Force – Example 1 Using Weight F = ma 35 N – N = (5.0 kg)a N = (5.0 kg)a a = m/s/s TOC 5.0 kg 35 N What is the acceleration?

2 Net Force – Example 2 Using Weight F = ma F – N = (8.0 kg)(+3.5 m/s/s) F = N TOC 8.0 kg F = ? What is the force? (Accelerating upwards at 3.5 m/s/s)

3 Whiteboards: Using Weight 11 | 2 | 3 | 4 | TOC

4 +2.69 m/s/s W 8.00 kg 100. N F = ma, weight = (8.0 kg)(9.81 N/kg) = N down Making up + = (8.0kg)a N = (8.0kg)a a = 2.69 m/s/s Find the acceleration:

5 -1.81 m/s/s W 15.0 kg 120. N F = ma, wt = (15.0 kg)(9.81 N/kg) = N down = (15.0kg)a N = (15.0kg)a a = m/s/s It accelerates down Find the acceleration:

6 +180 N W 16 kg F F = ma, wt = (16 kg)(9.81 N/kg) = N down = (16.0 kg)(+1.5 m/s/s) F – N = 24 N F = N ≈ 180 N Find the force: a = 1.5 m/s/s (upward)

7 +637 N W 120. kg F F = ma, wt = N downward = (120. kg)(-4.50 m/s/s) F – N = -540 N F = N ≈ 637 N Find the force: a = m/s/s (DOWNWARD)

8 +16,900 N W 120. kg F First, suvat: s = m, u = m/s, v = 0, a = ? use v 2 = u 2 + 2as, a = m/s/s F = ma, wt = N downward = (120. kg)( m/s/s) F – N = N F = ≈ 16,900 N 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?

9 -39.3 m W 2.10 kg 14.0 N If this box is initially at rest, what is its displacement in 5.00 seconds?

10 +1640 N W 52.0 kg F F = ma, a = m/s/s (from kinematics) wt = N downward = (52.0 kg)( m/s/s) F = N 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?

11 +62.2 N W 65.0 kg F F = ma, a = m/s/s (from kinematics) wt = downward = (65.0 kg)( m/s/s) F = N 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?

12 W F F = ma, wt = 1176 N downward = (120. kg)(-4.50 m/s/s) F – N = -540 N F = N Find the force: a = m/s/s (DOWNWARD) 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 120. kg

13 1.57 kg W m 13.6 N F = ma, wt = m(9.81 m/s/s) downward = 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 = kg ≈ 1.57 kg Find the mass: a = 1.12 m/s/s (downward)


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