(Mon) A 75 kg man jumps off of the roof of his garage 8 m above the ground and lands on his trampoli...
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At equilibrium: F net = 0 PowerPoint PPT Presentation


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(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).

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At equilibrium: F net = 0

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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)


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)

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


At equilibrium f net 0

(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)


At equilibrium f net 0

(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)


At equilibrium f net 0

(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)


At equilibrium f net 0

(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


At equilibrium f net 0

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