Terminal Speed for Skier Going Down Slope

  • Thread starter adamwest
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  • #1
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Homework Statement



What is the terminal speed for an 75.0kg skier going down a 35.0∘ snow-covered slope on wooden skis μk= 0.060?

Assume that the skier is 1.60m tall and 0.300m wide.

Express your answer using two significant figures with the appropriate units.

Area skier = A = (1.60m x 0.300m) = .48 m^2

Homework Equations



D = (1/4)A*v^2
Fnet = D

The Attempt at a Solution



(mgsinθ) - (μk*mgcosθ) = (1/4)A*v^2

=> vterminal = sqrt((4((mgsinθ) - (μk*mgcosθ))) /A )

vterminal = sqrt((4((75.0 * 9.80 * sin(35)) - (.060 * 75.0 * 9.80 * cos(35))))/.48)

vterminal = 56.68 m/s = 57 m/s (rounded to 2 sig figs)

Mastering Physics says that this answer is wrong but I cannot find my error. I have checked over the problem and my solution about a dozen times. I figure more experienced eyes may help find my ruinous mistake. Thank you! :)
 

Answers and Replies

  • #2
TSny
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Your work looks good to me. The data in the problem is given to 3 significant figures. Maybe you are supposed to keep 3 sig figs in the answer. Don't know.
 
  • #3
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Thanks for double checking for me, I really appreciate it. :)

I actually tried submitting it with 3 sig figs as well but it said it was wrong. I am going to email my professor and see if they can take a look.
 
  • #4
PhanthomJay
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D = (1/4)A*v^2
I believe this equation for drag force assumes a spherical shape for the drag coefficient (C_drag = about 0.5) . If the skier is standing more or less upright rather tham curled up like a ball, the drag coefficient is higher, and thus the terminal speed will be less. The quadratic drag force is actually 1/2(C_drag)ρ(A)v^2, where ρ is the air density(about 1.2 kg/m^3), and C_drag for a flat exposed surface might be more like 1.0 or so.
When I last went snow tubing, I knew my speed was much faster when lying back instead of sitting up. And faster than that when going with my grandkids 2 or more per tube and faster than that with several tube loads of kids hooked together in a lying down position! What a rush!
 
  • #5
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It looks like you were right, PhantomJay. I should have seen this before but I didn't check my units. The answer is as follows:

sqrt((2mg/ACp)*(sinθ-μkcosθ))

If you do it this way then you get all of the correct units (m/s) in the end after taking the square root. I used the following:

p (rho) = 1.2 kg/m^3
m = 75.0 kg
g = 9.80 m/s^2
θ = 35 degrees
μk = 0.060
C = 0.8

When you plug all of these into the equation you get a terminal velocity of v = 41 m/s (to 2 significant figures), which is the correct answer.

The only thing I am still unclear on is why we use a drag co-efficient of C = 0.8, which closely resembles a cylinder falling end-down, instead of a drag co-efficient of C = 1.1, which closely resembles a cyclinder falling side-down. I feel like a skier standing upright is affected by drag more similarly to the side of a cylinder than the bottom of one. Perhaps it has to do with the fact that they are falling at an angle? I don't know.

Anyways, thank you all for your help! :)
 
  • #6
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Thank you So much Sir/Madam, i tried a thousand times, spent an hour and 15 mins trying, but failing. Thank you again!
 

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