Question

8. a. What is the kinetic energy of a 1000 kg mystery machine van travelling at 15 m/s? Givens:
formula:
b. How much work has the van done? (hint:Energy to Work)
c. How much power is Freddy's beloved van putting out if it reaches this speed in 3 seconds?
Givens:
formula:
9a. The Wolfman’s kinetic energy is 1000 Joules as he speeds up to 5m/s, what is his mass?
Givens:
formula:
b. What is the Wolfman’s horizontal applied Force when he slams into Scooby and Shag and comes to rest after 10m? Givens: hint: you can’t use F=mg, the object isn’t falling, recall work =KE)
10a. A 60 kg Scooby is held 3.6m above a pool magical elixir. What is Scooby’s potential energy?
Givens:
formula:
b) If Scooby fell, What would be his kinetic energy right before he hit the pool if he had a velocity of 8.4 m/s?
c) How much work is occurring at that moment?
e) If Scooby hits the liquid with the power of 95 Watts how long was he in free fall?
Givens:
formula:

Answer 1

8a.
Givens: m = 1000 kg, v = 15 m/s
Formula: KE = 1/2 m v^2
KE = 0.5(1000)(15^2) = 0.5(1000)(225) = 112,500 J

8b.
Work done = change in kinetic energy (assuming start from rest) = 112,500 J

8c.
Givens: Δt = 3 s
Formula: Power = Work / time
Power = 112,500 J / 3 s = 37,500 W

9a.
Givens: KE = 1000 J, v = 5 m/s
Formula: KE = 1/2 m v^2 ⇒ m = 2·KE / v^2
m = 2(1000) / (5^2) = 2000 / 25 = 80 kg

9b.
Givens: initial KE = 1000 J, stops over d = 10 m
Use work-energy: Work by stopping force = ΔKE = -1000 J
Work = F·d ⇒ F = Work / d = -1000 J / 10 m = -100 N
Magnitude of the (average) horizontal stopping force = 100 N (directed opposite the motion)

10a.
Givens: m = 60 kg, h = 3.6 m, g = 9.8 m/s^2
Formula: PE = m g h
PE = 60·9.8·3.6 = 2,116.8 J ≈ 2.12 × 10^3 J

10b.
Givens: m = 60 kg, v = 8.4 m/s
Formula: KE = 1/2 m v^2
KE = 0.5·60·(8.4^2) = 30·70.56 = 2,116.8 J (same as PE)

10c.
The work done by gravity in the fall (conversion of PE to KE) = 2,116.8 J

10e.
If you interpret "hits the liquid with the power of 95 W" as average power P = energy / time, and take the relevant energy = 2,116.8 J, then
time = energy / power = 2,116.8 J / 95 W = 22.28 s.

Note: that 22.3 s is not the actual free-fall time from 3.6 m. The true free-fall time is
t = sqrt(2h/g) = sqrt(2·3.6 / 9.8) ≈ 0.857 s.
If the fall lasted 0.857 s, the average power during the fall would be 2,116.8 J / 0.857 s ≈ 2,470 W.