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The solution is on the next slides. It will be removed Mon AM

The solution is on the next slides. It will be removed Mon AM. Determine the “ideal” speed (mi/h) of the car at points 2, 3, and 4 as it travels the length of the coaster. The car starts from rest at point 1 which is 50.00 m higher than point 4. Point 2 is 5.00 m higher than point 4.

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The solution is on the next slides. It will be removed Mon AM

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  1. The solution is on the next slides. It will be removed Mon AM • Determine the “ideal” speed (mi/h) of the car at points 2, 3, and 4 as it travels the length of the coaster. • The car starts from rest at point 1 which is 50.00 m higher than point 4. • Point 2 is 5.00 m higher than point 4. • Point 3 is 30.00 m higher than point 4.

  2. Knowns: KE1 = 0 ; PE4 = 0 ; h1 = 50.00 m ; h2 = 5.00 m h3 = 30.00 m TE1= TE2 = TE3 = TE 4 h4 = 0 m ; v1 = 0 mi/h Unknowns: v2, v3, & v4 (mi/h) Equations: KE = 0.5mv2 ; PE = mgh ; TE = PE +KE

  3. Solve/Substitution: No, you were not given the mass. So set everything up to solve as if you had the mass, but leave the symbol, m, in place of the mass. This is an ideal situation, no loss of energy due to friction or anything else. So the total energy (TE) remains constant throughout. First write out all of the equations for the TE at each point: TE1= KE1 + PE1 ;TE2= KE2 + PE2;TE3= KE3 + PE3;TE4= KE4 + PE4 Now work with points one and two: TE1=TE2 KE1 + PE1 = KE2 + PE2 0.5mv12 + mgh1=0.5mv22 + mgh2 Notice that the mass shows up in each term, therefore it cancels out, you do not need it. Also, v1 = 0: 0.5mv12 + mgh1=0.5mv22 + mgh2 gh1=0.5v22 + gh2 Now rearrange to solve for v2: 0.5v22=gh1- gh2 v22= 2(gh1- gh2) v22= 2g(h1- h2) 0

  4. Solving Continued Now you can substitute in the actual values and calculate the speed: Follow the same process to solve for the speeds at points 3 and 4 v3 = 44.3 mi/hv4 = 70.0 mi/h

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