/20 marks Name: Module 6: Exemplary Level (2 marks) 8) A 60.0 kg student starts from rest (Point A) and slides 26.0 m down a hill where the average force of friction is 55.0 N. What is her final speed at the bottom of the hill (Point B) if the vertical drop is 17.0 m? TO 26.0 m 17.0 m B m = 60.0kg Ff = 5 5. ON Box 10f maboy V2= 285. 53 vd = 17. Om 2 = 16.9 m/s d = 26.0 m AEK + AEP = W ( muz _ mv . 2" ) + ( might 0 ) = Fxd intv = 0 2 - mgh = F d 2 mva = Fdtugh 2 v2 2 = V2 ( mgh + Fd ) F=mn due to friction V2 = V2 ligh - mind) V2 = 750 (60. 9.81-17- 35. 26) v2 = 12 - 8566