EXAMPLE 5.4Wax Your SkisGOAL Calculate the change in gravitational potentialenergy for different choices of reference level.PROBLEM A 60.0 kg skier is at the top of a slope, asshown in the figure. At the initial point , she is 10.0 mvertically above point B. (a) Setting the zero level forgravitational potential energy at B, find the gravitationalpotential energy of this system when the skier is at 4 andthen at B. Finally, find the change in potential energy of theskier-Earth system as the skier goes from point to pointB. (b) Repeat this problem with the zero level at point4. Repeat again, with the zero level 2.00 m higher thanpoint Ⓡ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 PE₁. The location chosen for y = 0 is also thezero point for the potential energy, because PE = mgy.SOLUTION(A) Let yenergy.Find PE, the potential energy at 4.=0 at B. Calculate the potential energy at A and at B, and calculate the change in potentialPE= = 0 at by choice. Find thedifference in potential energy betweenA and ⓇPE; = mgyi5.88 × 10³ J=10.0 m(60.0 kg)(9.80 m/s²)(10.0 m) =PE - PE; = 0 - 5.88 × 10³ J = -5.88 × 10³ J PRACTICE ITUse the worked example above to help you solve this problem. A 54.0 kg skier is at the top of a slope, asshown in the figure. At the initial point 4, she is 8.0 m vertically above point Ⓡ.(a) Setting the zero level for gravitational potential energy at B, find the gravitational potentialenergy of this system when the skier is at A and then at B. Finally, find the change in potentialenergy of the skier-Earth system as the skier goes from point to point Ⓡ.PE; =PE =APE =(b) Repeat this problem with the zero-level at point 4.PE;PEf =ΔΡΕ ===JJ(c) Repeat again, with the zero level 2.00 m higher than point Ⓡ.PE;JPEfAPE ==JJJ

At the the point A, gravitational potential energy is zero. At any height y the gravitational potential is PE = mgy. Where, m is mass of the skier = 60 kg. g = 9.8 m/s2 is gravitational acceleration. (a) At A, y = yi = 10.0 m. The potential energy at A, PEi = mgyi = (60.0 kg) (9.80 m/s2) (10.0 m) = 5.88 x 103 J At A, y = yi = 10.0 m. PEf = 0 at B by choice The change in potential energy ∆PE = PEf - PEi = - 5.88 x 103 J = - 5.88 x 103 J(b) Setting A as zero level. At A, y = yi = 0 m. PEi = 0 at A by choice. At B, y = yf = -10.0 m. The potential energy at B, PEf = mgyf = (60.0 kg) (9.80 m/s2) (- 10.0 m) = - 5.88 x 103 J The change in potential energy ∆PE = PEf - PEi = 0 - (- 5.88 x 103 J) = 5.88 x 103 J(c) Setting y = y0 = 2 m above B as zero level. At A, y = yi = 10.0 m. PEi = mg(yi - y0) = (60.0 kg) (9.80 m/s2) (10.0 m - 2.0 m) = 4.704 x 103 J. At B, y = yf = -2.0 m. The potential energy at B, PEf = mgyf = (60.0 kg) (9.80 m/s2) (- 2.0 m) = - 1.176x 103 J The change in potential energy ∆PE = PEf - PEi =-1.176 x 103 J - 4.704 x 103 J = - 5.88 x 103 J