A skier slides horizontally along the snow for a distance of 11.9 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0458. Initially, how fast was the skier going?
I do not know how to start this. Thank you.
Physics(Please help) - Elena, Saturday, June 2, 2012 at 8:59am
ΔKE =KE2 –KE1 = 0 -m•v²/2.
W(fr) =μ•m•g•s•cosα,
where α is the angle between the friction force and displacement.
α =180º, cos α = -1.
-m•v²/2 = - μ•m•g•s,
v = sqrt(2•μ• g•s).
So for v would I do sqrt(2*11.9*9.8*0.0458) ?
3 answers
Yes
I got 10.68 as my answer but my homework said this was incorrect.Why?
v = sqrt(2•μ• g•s)= sqrt(2•0.0458•9.8•11.9) = 3.27 m/s