Question

A 62.5 kg skier is moving at 6.15 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00 m long. The coefficient of kinetic friction between this patch and her skis is 0.320. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.35 m high.

A)How fast is the skier moving when she gets to the bottom of the hill?

BHow much internal energy was generated in crossing the rough patch?

Answer 1

mass of skier= 62.5 kg

velocity=6.15 m/s

length=4.00 m

coefficient of kinetic friction between this patch and her skis = 0.320.

hill height= 3.35 m

Let us use the energy equations

Kinetic energy,E=1/2mv²

where, mass= m

velocity= v

E=F(d) which is a force F over a distance d =4 m

E=mgh for a mass in gravity falling a distance h=3.35m

_k =Fk/Fn or coefficient of friction is ratio of Frictional force to normal force (mg)

so equation is inital energy of velocity (v1) minus friction energy plus energy of drop = final energy of velocity (v2)

1/2mv1² - mg(k)(D) + mgh = 1/2 mv₂²

divide by m on both sides

1/2 v1² - g(k)(D) + gh = 1/2v2²

multiply by 2 on both sides

v1² + 2g(h- (k)(D)) = v2²

v2 = sqrt(v1² + 2g(h- (k)(D)))

so substitute the values

v2 = sqrt (6.15² + 2(9.81)(3.35-0.320(4)))

v2 = sqrt 78.39

v2 = 8.854 m/s

No energy is created or destroyed, but energy of velocity was converted to heat in the rough patch in the amount of F(D) joules

E =mg(k)(D)

E = 62.5(9.81)(.32)(4)= 784 J