Your firm is designing a new roller-coaster have a total mass (including passeng
ID: 1581973 • Letter: Y
Question
Your firm is designing a new roller-coaster have a total mass (including passengers) of 800 kg and will travel freely along the winding frictionless track shown in the figure below ride. The permit process requires the calculation of forces and accelerations at various i important locations on the ride. Each roller-coaster car wll Points A, E, and G are on horizontal straight sections, same helght of 10 m above the ground. Point c is at a helight of 10 m above the ground on an inclined section of slope angle 30. Point B is at the crest of a hl ground level at the bottom of a valley; the radius of curvature the same helight as points A, E, and G. At point A the speed of the car is 22.9 m/s at both of these points is 20 m. Point F is at the middle of a banked horizontal curve with a radlus of (a) If the car is just barely to make it over the hill at point B, what must be the height of point B above the ground? (b) If the car is to just barely make it over the hill at point B, what should be the magnitude of the force exerted by the track on the car at that point? kN (c) What will be the acceleration of the car at point C? (d) what will be the magnitude and direction of the force exerted by the track on the car at point D? kN directed (e) What will be the magnitude and direction of the force exerted by the track on the car at point F? X KN directed upward from the horiontal. (n At point G a constant braking force is to be applied to the car, bringing it to a hait in a distance of 20 m. What is the magnitude of this required braking force Choose the system to include Earth, the track, and the car. Then there are no external forces to do work on the system and change its energy. Use Newton's second low and the work energy theorem to describe tho system's energy transformations to point G . and then the work.energy theorem with friction eo determine the biking eree that benge te ano Draw free-body diagrams of the cart at each relevant pointExplanation / Answer
Mass of the car m =800 kg
initial speed at A is V_A =22.9 m/s
A)
from law of conservation of energy
mgh_B =(1/2) m(V_A)^2
h_B =(V_A)^2/2g
h_B =26.73 m
net height from the ground H_B =10 m+26.73 m =36.73m
B)
Force exerted by track on car
W =mg =7.840*10^3 N
C)
The acceleration at point C
a_c =g sin(theta) =(9.8 m/s^2) sin30 =4.9 m/s^2
D)
total length BD =(H_B+SR)/sin30 =(36.73 m+20 m)/0.5 =113.46 m
let V is the velocity at point D
apply equation of motion due to acceleration
v^2 =0+2(g sin30)*BD = 2*9.8*113.46
V^2 = 2223.816 (m/s)^2
external force F_D = (mV^2/R)-mg =81.1126*10^3 N
The direction of this force is upwards.
E)here r =35 m
horizontal force F_H =m(V_A)^2/r =800*22.92/35== 11.9865*10^3 N
and F_v =mg =800*9.8=7.84*10^3 N
F_F =sq.root [(F_v)^2+(F_H)^2]
=14.322*10^3 N
direction ? =tan^-1(F_v/F_H)=56.8125degrees
F)
from conservation of energy
F_G*S =1/2(m)(V_A)^2
net force at G is
F_G =1/2[(m)(V_A)^2]/S
where S =20 m
substitute yhe given data in above equation we get
F_G = 10.488*10^3 N
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