A 800 kg roller-coaster train is initially at the top of a rise, at point A. It

Question

A 800 kg roller-coaster train is initially at the top of a rise, at point A. It then moves 155 ft, at an angle of 40.0° below the horizontal, to a lower point B.
(a) Choose the train at point B to be the zero configuration for gravitational potential energy. Find the potential energy of the roller coaster-Earth system when the train is at points A and B, and the change in potential energy as the coaster moves.

potential energy
______ J at point A
______ J at point B
change in potential energy (UB - UA)
______ J

(b) Repeat part (a), setting the zero configuration when the train is at point A.
potential energy
______ J at point A
______ J at point B
change in potential energy (UB - UA)
______ J

Explanation / Answer

A) Now 1 ft = 0.3048 meter 155 ft = .3048*155 = 47.244 m We knowpotential energy is equal to: Ue =mgh In thiscase, h = 47.244m*sin(40) = 30.36 m Potential energy atA: Ue =800kg*9.81m/s2*30.36m = 238264.28 J Potential energy atB: Ue =800kg*9.81m/s2*0 = 0J Difference inpotential energy: ?Ue =238264.28 J -0J = 238264.28 J B) h2 =this time choosing y = 0 atpoint , which is another down the same slope from point Potential atA: Ue =mg(h+h2) = 800kg*9.81m/s2*(30.36+h2]J Potential atB: Ue =mgh2 = 800kg*9.81m/s2*h2 J Difference inpotential: ?Ue =800kg*9.81m/s2*(30.36+h2]J - 800kg*9.81m/s2*h2 = 800*9.81*30.36 = 238265.28 J

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