Conservation of Energy and Non Conservative Force, Skier down slope with Friction
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Conservation of Energy and Non Conservative Force, Skier down slope with Friction

[From: ] [author: ] [Date: 12-03-31] [Hit: ]
8N/kg)(50m) = (60v^2)/2 + (6.v = 27.......
I've been stuck on one problem that I haven't quite figured out how I can approach to solve.

"A 60kg Skier starts from rest at the top of a 50m high slope. If the work done by friction is -6.0x10^3 J, what is the speed of the skier on reaching the bottom of the slope?"

Lots of things confuse me here, starting with "work" done by friction, as well as no delta X, only a height of y. How can I solve this?

Thank you.

-
work done by friction is just thermal energy that friction causes

Initial energy equals finals energy

initial Energy = final energy

Initial energy is only gravitational potential energy since the skier is not moving
At the bottom of the hill, the potential energy converts to kinetic and thermal due to friction

Eg = Ek + Eth
mgh = (mv^2)/2 + (6.3x10^3)
(60kg)(9.8N/kg)(50m) = (60v^2)/2 + (6.3x10^3)
(60v^2)/2 = 29400 - 6300
v = sqrt((2 x 23100)/60)
v = 27.8m/s
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