Question

A skier weighing 0.80KN comes down a frictionless ski run that is circular R=30m at the bottom, as shown.if her speed is 12m/s at point A ,what is her speed at the bottom of the hill(pointB)?

Answer 1

calculate the distance from point A to point B

that is D= 30(1 - cos(40)) = 7.01 m

now, mechanical energy that is KE + PE is conserved

hence, initial mechanical energy = final mechanical energy -----------------(A)

also, mass = 0.80*1000/9.8 = 81.63 kg

also PE = mgd and KE = .5mv²

now from equation A, LHS

initial mechanical energy = (.5 * 81.63 * 12²) + (800 * 7.01)

initial mechanical energy = 11492.29 J

also RHS,

final mechanical energy = 0.5*81.63*v² + m*g*0 { final distance d =0 }

fianl mechanical energy = 0.5*81.63*v²

now equating initial and final mechnical energy, we get

v² = 281.57

v = 16.78 m/s