In the figure below, a single frictionless roller-coaster car of mass m = 621 kg
ID: 1438896 • Letter: I
Question
In the figure below, a single frictionless roller-coaster car of mass m = 621 kg tops the first hill with speed v_0 = 16.2 m/s at height h = 57.0 m. (a) How much work does the gravitational force do on the car from the initial point to point A? J (b) How much work does the gravitational force do on the car from the initial point to point B? J (c) How much work does the gravitational force do on the car from the initial point to point C? J (d) If the gravitational potential energy of the car-Earth system is taken to be zero at point C, what is its value when the car is at B? J (e) If the gravitational potential energy of the car-Earth system is taken to be zero at point C, what is its value when the car is at A? J (f) If mass m were doubled, would the change in the gravitational potential energy of the system between points A and B increase, decrease, or remain the increase decreaseExplanation / Answer
W = mgh. (m = mass, g = gravity and h = height)
m= 621 kg
g= 9.81 m/sec^2
h= 57 m, V0 = 16.2 m/sec
a) Point A
W = W1 - W2
W = mg(h-h) = 0
W = 0 J
b) Point B
W = mg(h - h/2) = 621*9.81*(57 - (57/2))= 1.74*10^5
W = 1.74*10^5 J
c) Point C
W= mg(h - 0) = 621*9.81*57 = 3.47*10^5
W = 3.47*10^5 J
PE = mgh
d) PEb = mgh/2 = 621*9.81*57/2 = 1.74*10^5
PEb = 1.74*10^5 J
e) PEa = mgh = 621*9.81*57 = 3.47*10^5
PEa = 3.47*10^5 J
f) Increase, since the mass is proportional to PE, change in the gravitational potential energy also doubles.