LEARN MDRE REMARKS These calculations ShOW that the change in the gravitational potential energy when the skier goes from the top of the slope to the bottom is -5.88 X 103 J, regardfess ofthe zero i'evei selected. QUESTION If the angle of the slope is increased, the change of gravitational potential energy between the two heights: {Select all that apply.) I:l increases. l:l remains the same. D depends only on the difference between the two heights. D depends on the path followed. I:l decreases. l:l sometimes decreases and sometimes increases. PRACTICE IT Use the worked example above to help you solve this problem. A 151.0 kg skier is at the top of a slope, as shown in the figure. At the initial point @, she is 12.1] m vertically above point @. {3] Setting the zero level for gravitational potential energy at @, find the gravitational potential energy of this system when the skier is at and then at @. Finally, find the change in potential energy of the skier Ea rth system as the skier goes from point @to point @. f (b) Repeat this problem with the zerolevel at point @. f (c) Repeat again, with the zero level 2.00 rn higher than point @. EXERCISE HINTS: GETTING STARTED | I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. If the zero level for the gravitational potential energy is selected to be midway down the slope, 6.00 m above point , find the initial potential energy, the final potential energy, and the change in potential energy as the skier goes from point A to B in the figure. initial kJ final kJ change KJ10.0 m B) GOAL Calculate the change in gravitational potential energy for different choices of reference level. PROBLEM A 60.0 kg skier is at the top of a slope, as shown in the figure. At the initial point @, she is 10.0 m vertically above point . (a) Setting the zero level for gravitational potential energy at B, find the gravitational potential energy of this system when the skier is at @ and then at B. Finally, find the change in potential energy of the skier-Earth system as the skier goes from point & to point B. (b) Repeat this problem with the zero level at point @. Repeat again, with the zero level 2.00 m higher than point B. STRATEGY Follow the definition and be careful with signs. @ is the initial point, with gravitational potential energy PE, and @ is the final point, with gravitational potential energy PE. The location chosen for y = 0 is also the zero point for the potential energy, because PE = may. SOLUTION (A) Let y = 0 at B. Calculate the potential energy at @ and at B, and calculate the change in potential energy. Find PE , the potential energy at A. PE, = mgy; = (60.0 kg)(9.80 m/s )(10.0 m) = 5.88 x 103 J PE = 0 at B by choice. Find the PE - PE, = 0 - 5.88 x 10' ] = -5.88 x 103 ] difference in potential energy between and B(B) Repeat the problem if y = 0 at @, the new reference point, so that PE = 0 at Find PE , noting that point is now at PE = may, = (60.0 kg)(9.80 m/s )(-10.0 m) = y = -10.0 m. -5.88 x 10 J PE, - PE. = -5.88 x 10' J - 0 = -5.88 x 10* ] (C) Repeat the problem, if y = 0 two meters above B Find PE , the potential energy at A PE, = mgy; = (60.0 kg)(9.80 m/s )(8.00 m) = 4.70 x 10 J Find PE the potential energy at PE = mgy, = (60.0 kg)(9.80 m/s )(-2.00 m) = -1.18 x 10 J Compute the change in potential PE - PE. = -1.18 x 10J - 4.70 x 10# J = energy. -5.88 x 10' J( b) The potential energy at A [yi = 0] P.E; = may. JA (o ,0 ) = 610 X 9.80 X O 72 PEi = 0 The potential energy at B [ ys = - 12.0 m ] $12.0m PEf = 61.0 X 9.80 x ( - 12.0) PE! = - 7.17 x 10# J The change in potential energy from A to B B (1 - 12. 0) APE = PEJ - PE: = - 7017 X 10 - 0 APE = - 7017 X 10 J A( , 10.0) ( C) The potential energy at A [ yi = 10.0m] PE; = may ; = 61 0 X 9. 80 x 10.0 12.0m PE: = 597 8 J = 5. 98 x 10' J (0, 0) -7 x The potential energy at B [ Y, = - 2.0 m ] PEJ = may, B (2 , - 2.0 ) = 61.0 x 9.80 x ( - 2.0 ) PEJ = - 1195. 6J = - 1 . 19 x10 JThe change in potential energy from A to D APE = PE, - PEi = - 1195.6 - 597.8 = - 7173. 6 APE = - 7. 17 X 10' J A ( 2: , 6.0 ) The potential energy at A [ yi = 6.0m ) PE: = may: = 610 X 9.80 X 6.0 12.0m ( 0, 0 ) 7x PE: = 3586. 8 J = 3. 59 x10 J The potential energy at B [ Y,=- 60m) PE = mgys B(X - 6. 0) = 610 X 9.80 x (-6.0) PE = - 3586.8 J =- 3- 59 x 10' J The change in potential energy from A to B APE = PEf - PE: = - 3586.8- 3586.8 = - 7173.6 APE = - 7,17 X 10 J