In the figure below, a frictionless roller coaster of mass m = 533 kg tops the f
ID: 2229541 • Letter: I
Question
In the figure below, a frictionless roller coaster of mass m = 533 kg tops the first hill with speed v0 = 4.7 m/s at height h = 47 m.
(a) How much work does its weight do on it from the initial point to pointA?
(b) How much work does its weight do on it from the initial point to pointB?
(c) How much work does its weight do on it from the initial point to pointC?
(d) The gravitational potential energy of the coaster-Earth system is taken to be zero at pointC. What is its value when the coaster is at pointB?
(e) What is its value when the coaster is at pointA?
(f) If massmwere doubled, would the change in the gravitational potential energy of the system between pointsAandBincrease, decrease, or remain same?
Explanation / Answer
a)since the energy at both the points are the same , so the weight done by the weight is 0 b)work done=loss in potential energy =mgh-mgh/2 =mgh/2 c)work done=loss in potential energy=mgh-0 =mgh d)applying law of conservation of energy, energy at b=mgh/2 e)it is mgh according to law of conservation of energy. f)it would increase and become mgh