P. O. D.19, PAGE 2 NAME POD 19-Prob#5) Roller Coaster of Love A frictionless rol
ID: 1584654 • Letter: P
Question
P. O. D.19, PAGE 2 NAME POD 19-Prob#5) Roller Coaster of Love A frictionless roller coaster car of mass m -825 kg tops the first hill with a speed vo-17.0 m/s at a height of 42.0m. How much does the gravitational force do on the car from the starting point to a. Point A? b. Point B? c. Point C? 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 d. Point B? e. Point A? f. If the mass of the car were doubled, would the change in the gravitational potential energy of the system between Point A and Point B increase, decrease, or remain the same?Explanation / Answer
a) Workdone by gravity from prom starting point to point A,
WA = m*g*(hi - hA)
= 0
b) Workdone by gravity from prom starting point to point B,
WB = m*g*(hi - hB)
= m*g*(h - h/2)
= m*g*h/2
= 825*9.8*42/2
= 1.70*10^5 J
c) Workdone by gravity from prom starting point to point C,
WC = m*g*(hi - hC)
= m*g*(h - 0)
= m*g*h
= 825*9.8*42
= 3.40*10^5 J
d) at point B, UB = m*g*hB
= m*g*h/2
= 825*9.8*42/2
= 1.70*10^5 J
e) at point A, UA = m*g*hA
= m*g*h
= 825*9.8*42
= 3.40*10^5 J
f) increases.
Because, if m is doubled U also becomes double.
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