A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snow-covered

Question

A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.50 m long.  The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.

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

v = ?  m/s

2.) How much internal energy was generated in crossing the rough patch?

E = ? J

I have no idea how to do this. Please show step by step with correct answers so that I can try to understand.

THANK YOU!

Explanation / Answer

we will use energy equations
E=1/2mv^2 for energy of a mass m at velocity v
E=F(d) which is a force F over a distance d =3.5m
E=mgh for a mass in gravity falling a distance h=2.5m

uk =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)

1/2mv1^2 - mg(uk)(D) + mgh = 1/2 mv2^2

divide by m

1/2 v1^2 - g(uk)(D) + gh = 1/2v2^2

multiply by 2
v1^2 + 2g(h- (uk)(D)) = v2^2
v2 = sqrt(v1^2 + 2g(h- (uk)(D)))

so plug in what we know

v2 = sqrt (6.5^2 + 2(9.81)(2.5-0.3(3.5)))
v2 = sqrt 70.7
v2 = 8.4 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(uk)(D)
E = 62(9.81)(.3)(3.5)= 638.6 J

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