a frictionless roller coaster of mass m = 615 kg tops the first hill with speedv

Question

a frictionless roller coaster of mass m = 615 kg tops the first hill with speedv1 = 5.9 m/s. Assumethat the first hill is h = 59 m tall. Figure 10-28(a) How much work does the gravitational force do on it from theinitial point to point A? 1 J (b) How much work does the gravitational force do on it from theinitial point to point B? 2 J (c) How much work does the gravitational force do on it from theinitial point to point C? 3 J (d) If the gravitational potential energy of the coaster-Earthsystem is taken to be zero at point C, what is its valuewhen the coaster is at point B? 4 J (e) If the gravitational potential energy of the coaster-Earthsystem is taken to be zero at point C, what is its valuewhen the coaster is at point A? 5 J (f) If mass m were doubled, would thechange in the gravitational potential energy of the system betweenpoints A and B increase, decrease, or remainsame? remain same decrease increase

Explanation / Answer

Alright I figured it out. . W = U or -(Uf -Ui) Ui = mgh a) At point A (its at rest....) W = U Ui = mgh Uf = mgh W = (Uf Ui ) = (mgh mgh) = 0 b) At point B W = U Ui = mgh Uf = mgh2 W = (Uf Ui ) = (mg(h/2) mgh) = mgh2 = 615kg 9.8 (59m/2) =177796.5J c) At point C W = U Ui = mgh Uf = mg 0 W = (Uf Ui ) = (0 mgh) = mgh = 615 9.8 59m = 355593J d) The gravitational potential energy at B is UB = mgh2 = 615 9.8 (59/2) =177796.5J e) The gravitational potential at A is UA = mgh = 615 9.8 59m = 355593J f) If you double the mass, all results double.

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