Phy410 Homework -Ch 07 begins with a \"catapult launch\" in which the train is p
ID: 2034105 • Letter: P
Question
Phy410 Homework -Ch 07 begins with a "catapult launch" in which the train is propelled by "linear synchronous motors" from 0 to 25 m/s (-55 mph) along 66 m of horizontal track. The train itself has a mass of approximately 5000 kg, and holds 24 people of 75 kg each on average. a) Determine the average propelling force (in Newtons, and then lbs) for a train with 24 people. 7) California Screamin' in Disney's California Adventure park is one of my favorite roller coasters. I Assume the force is horizontal, along the track. Ignore friction.) Use WET (KE&PE;). b) Determine the time it takes (assuming constant force and acceleration). c) Determine the average power needed to launch the train, first in watts, then in h.p. Near the end of the ride, the train goes through a "Teardrop" loop, as shown in the diagram that follows. Point C is about 20 m above point A, where the train car enters at about 22 m/s (-50 mph). Point B which the train car's velocity is vertically up, is at a height of about 15 m. The radius r for the curve from B to C is about 5.0 m. A 75 kg man sits in the car. Ignore the size of the train car itself, and ignore friction. Assume the only force on the man by the car is the normal force from the seat on his rear-no friction, and no force on his back. 20 m 15 m A 22 m/s (More questions on next page)Explanation / Answer
7.
a)
vi = initial velocity = 0 m/s
vf = final velocity = 25 m/s
d = distance travelled = 66 m
a = acceleration = ?
Using the equation
vf2 = vi2 + 2 a d
252 = 02 + 2 a (66)
a = 4.73 m/s2
M = mass of train + mass of 24 people = 5000 + 24 (75) = 6800 kg
average force is given as
F = M a = 6800 x 4.73 = 32164 N
b)
using the equation
vf = vi + at
25 = 0 + 4.73 t
t = 5.3 sec
c)
P = power = F d / t = 32164 x 66/5.3 = 400532.83 watt = 537 hp