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

Answer:

$1.51 * 10^4 N$

Explanation:

The following parameters are given

initial speed u = 22.8 m/s

final speed v = 2.9 m/s

distance s = 5.55 m

mass of car m = 328 kg

deceleration a= ??

Using Newton's third equation of motion

$v^2 = u^2 -2as$

Here, we chose -a since the car is decelerating (negative acceleration)

Hence $2.9^2 = 22.8^2 - (2*a*5.550)$

Solving the equation, we obtain a = 46.074 $m/s^2$

Hence the braking force = mass of car * deceleration

= 328 * 46.074

= $1.51 * 10^4 N$