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At equilibrium: F net = 0

(Mon) A 75 kg man jumps off of the roof of his garage 8 m above the ground and lands on his trampoline 1 m above the ground. If the total spring coefficient of the trampoline is 41 kN/m, how far does the trampoline flex down once he stops bouncing around? (5 min/5 pts).

candice-marsh
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At equilibrium: F net = 0

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  1. (Mon) A 75 kg man jumps off of the roof of his garage 8 m above the ground and lands on his trampoline 1 m above the ground. If the total spring coefficient of the trampoline is 41 kN/m, how far does the trampoline flex down once he stops bouncing around? (5 min/5 pts)

  2. (Mon) A 75 kg man jumps off of the roof of his garage 8 m above the ground and lands on his trampoline 1 m above the ground. If the total spring coefficient of the trampoline is 41 kN/m, how far does the trampoline flex down once he stops bouncing around? (5 min/5 pts) At equilibrium: Fnet = 0 8m ?m Fg = Fe or mg = kx 1m (75kg)(9.81m/s²) = (41kN/m)(x) 735.75 N = 41,000(x) N x = 735.75 / 41,000 m x = 0.0179 m Or 1.79 cm

  3. (Tue) A 2600 kg (5750lb) Escalade and a 1160 kg (2550lb) Cooper Mini are stopped at a stoplight. When the light turns green, and both accelerate to 20 m/s (45mph) in a distance of 300 m. How much less force did the Mini have to use to keep even with the Cadillac? (5 min / 5 pts)

  4. (Tue) A 2600 kg (5750lb) Escalade and a 1160 kg (2550lb) Cooper Mini are stopped at a stoplight. When the light turns green, and both accelerate to 20 m/s (45mph) in a distance of 300 m. How much less force did the Mini have to use to keep even with the Cadillac? (5 min / 5 pts) W = Fd W = ΔKE ΔF = 960 N W = KEf - KEi X Fd = ½mv² F = ½mv²/d ΔF = (½)(Δm)v²/d ΔF = (0.5)(2600kg-1160kg)(20m/s)²/(300m)

  5. (Wed) A 1000 kg elevator carries a max load of 800 kg. A constant friction force of 4000 N pulls down on the car as it goes up. What is the minimum power, in kW, that the motor must have to lift the car at 3 m/s? (10 min / 10 pts)

  6. (Wed) A 1000 kg elevator carries a max load of 800 kg. A constant friction force of 4000 N pulls down on the car as it goes up. What is the minimum power, in kW, that the motor must have to lift the car at 3 m/s? (10 min / 10 pts) Mass = Mass Car + Mass People Fpull Mass = 1000 kg + 800 kg = 1800 kg Constant speed means Net Force = 0 ΣFy = 0 or Fp = Ff + Fg Fp = 4000 N + (1800 kg)(9.81 m/s²) Fp = 21658 N Ff Fg P = W/t or P = Fv (use this one) P = (21658 N)(3 m/s) P = 64,974 W or…. P = 64.974 kW

  7. End of Week Procedures: • Add up all the points you got this week • Put the total number of points at the top of your page (out of 20 points) • List any dates you were absent (and why) • Make sure your name and period is on the top of the paper • Turn in your papers

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