Question

10. Assuming the skater is release from rest at point A which is 6.00 m above the ground, use the conservation of mechanical energy calculate the speed the of the skater at B, C, and D. (Show all work)

Answer 1

The working folmula here total mechanical energy is is the conserved. je, total mechanical energy = kinetic energy & potential energy (gravitational) = constant. Since initial energy at point a is Potential energy only (since it has been neleased from hest) hence total nechanical energy, E = mgha - Now, kinetic energy is kB=tmus a at potion point B is at point B potential energy UB= mgha hence , E = KB TUB 7 kB = E-UB - кra (hp - hо )

given, - hason e ho = 2n & m=75 kg hence, KBZ 75*9.80 (6-2) I 2 KB = 2940J 2. at B the potential energy is, UB= mgha 3) UB=(75*9.8*2) 2) UB= 1470 ] - 3. since, Imra= KB >> Vez Jooga to mis 25 =) VBZ 8.854377 448 mis VB = 8.85 mis lappoon

(4) similarly we can use Vet Wc = mgha a Kc = mgha-ve ke = my channel n n ke=7509.87 (6-4) * ke = 1470 I ħ at 6) potential energy c is The Ue = mghe = Ue = 75x9.8 * 4 ] 7 Ue = 2940J (6) The speed at point c is ves 23 mis 12x 1470 v 75 23 1970 mis » vez 6.260990337 mis =) Ve=6. 26 mis cappoon)

(7) again we will use, kot Up = mgha » ko = mgha-Up = ng Chanho) 2) ko= mg (ho-ho) ginen, ho=1.25m. hence, the kinetic energy at point - ko=75*9.8+(6-1-25) = ko = 3491.253 (8) The Potential energy at points Up = mgho → Vo = (75*9.8 x 1.25) 5 3) Up=918.753 9) the speed at point is Voz / 240 = 2 2x3491,25 mis =) = 9.648834126 m/s