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Energy & Non Conservative Forces (10 points)?
A skier of mass 116 kg starts from rest at the top of the ski slope of height 100 m shown in the figure below. If the total work done by friction is −4.5 x 10^4 J, what is the skier's speed when he reaches the bottom of the slope? (Figure not drawn to scale.)
http://www.webassign.net/giocp2/6-p-071.gif
Answer is 34.4 m/s
I need help getting there, please explain the work. Thank you.
2 Answers
- Mark PLv 78 years agoFavorite Answer
Initial potential energy = work done against friction + final kinetic energy
mgh = -4.5x10^4 + mv^2/2
(116)(9.8)(100) = 4.5x10^4 + (116)(v^2)/2
1.14x10^5 - 4.5x10^4 = 58v^2
v = 34.4 m/s
NOTE: The work done by friction is negative, the work done against friction is positive.
- 8 years ago
First you need to calculate the net work on the skier which is the work due to gravity plus the work due to friction:
gravitational work -> W_net=F*d = 9.81*M*d where M is the skiers mass and d is the height of the hill
W_net=1.14 x 10^5 J
Net work = 6.88 x 10^4 J
The net work is also his net kinetic energy at the bottom of the hill and since kinetic energy (KE) is
KE = W_net = 1/2*m*v^2
v = sqrt(2*W_net/m)
v = 34.4 m/s
