A 41.0-kg skier with an initial speed of 1.5 X 101 m/s coasts up a 2.50-m-high r
ID: 2039843 • Letter: A
Question
A 41.0-kg skier with an initial speed of 1.5 X 101 m/s coasts up a 2.50-m-high rise as shown below. The coefficient of friction between her skis and the snow is 0.0800. KE 2.5 m 35° a) Where do you define the gravitational potential energy Us to equal 0 J? b) If the skier has energy at the bottom of the hill state what kind it is and determine its value. c) If the skier reaches the top of the hill what kind of energy would she have? d) What affect does the friction have on her velocity at the top of the hill, if any? e) Explain what conservation of energy means and how it applies in this situation. BONUS: Determine the skier's velocity at the top of the hill if she makes it that far, or determine how high up the slope she goes before stopping. (10 points)Explanation / Answer
a)
at the bottom of hill
b)
the skier has kinetic energy at the bottom of hill
Ebottom = (1/2)*m*vbottom^2 = (1/2)*41*(1.5*10^1)^2 = 461.25
(c)
kinetic energy and gravitational potential energy
d)
friction decreases the velocity of the skier
(e)
conservation of energy
The total energy of a system remains constant
energy can neither be created nor destroyed , but can be changed from one form to another
there is no change in energy of the system
dE = 0
dUg + dKE + dEinternal = 0
dUg = cahnge in potential energy
dKE = change in kinetic energy
dEinternal = change in internal energy = f*L
L = length of the incline = 2.5/sintheta
f = u*m*g*costheta
Utop - Ubot + Ktop - Kbot + f*L = 0
m*g*h - 0 + (1/2)*m*vf^2 - (1/2)*m*vi^2 + u*m*g*costheta*h/sintheta = 0
gh + (1/2)*vf^2 - (1/2)*vi^2 + u*g*tantheta*h = 0
vf = sqrt( vi^2 - 2gh - 2*u*g*tantheta*h )
vf = sqrt(15^2 - (2*9.8*2.5) - (2*0.08*9.8*tan35*2.5))
vf = 13.16 m/s <<<<-------------ANSWER
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