Question

Dizzy is speeding along at 22.8 m/s as she approaches the level section of track near the loading dock of the whizzer roller coaster ride. a braking system abruptly brings the 328-kg car (rider mass included) to a speed of 2.9 m/s over a distance of 5.55 meters. determine the braking force applied to dizzy's car.

Answer 1

kinetic energy before braking = ½mv² = (½ * 328 *22.8²) = 85254 jouleskinetic energy after braking= ½mv² = (½ * 328 * 2.9²) = 1379 jouleschange in k.e. = (1379 - 85254 ) = -83874 joules[work energy theorem: work done = change in k.e.]work done by braking system= braking force * distance over which force acts.-83874 = -F * 5.55 m {F is in opposite direction to motion.}F = 83874 / 5.55F = 15,100 N

Answer 2

ANSWER
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change in car's speed = Vo - Vf = 22.8 - 2.9 = 19.9 m/s
distance = d = 5.55 m
t = time interval over which speed change occurs
a = deceleration of roller coaster car = 19.9/t

d = Vo(t) - 1/2at²
5.55 = 22.8t - 0.5at² = 22.8t - 9.95t = 12.85t
t = 5.55/12.85 = 0.43191 s
a = 19.9/0.43191 = 46.07 m/s² <= 46.07/9.81 = 4.7g {a rather high no. of "g" at the end of a roller coaster ride}
braking force = F = ma = 328(46.07) = 15,112 ≈ 15,100 N ANS

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