Question

Two children are sliding down a snowy hill which is inclined by
47.7 degrees with respect to horizontal. One has a rough sled which
has a coefficient of kinetic friction between it and the snow of
*μ**k*= 0.371; the other has a slippery sled which
has a coefficient of kinetic friction between it and the snow of
*μ**k* = 0.113. If both children start at the top
of the 48.1 m tall hill initially at rest, what is the difference
in speed between these two sleds at the bottom of the hill?

Answer #1

length of slope d = h/ sin(47.7) = 48.1/ sin(47.7) =65.032 m

apply work energy therom

Wgravity + W-friction = change in KE = 1/2*m(vf^2-vi^2)

for first child

mgh-μ *m*g*cos(47.7)*d = 0.5*m*vf^2

9.8*48.1 -0.371*9.8*cos(47.7)*65.032 = 0.5*vf^2

from here **vf = 24.99 m/s**

same apply for 2nd child with diffrent mu

9.8*48.1 -0.113*9.8*cos(47.7)*65.032 = 0.5*vf^2

**vf = 29 m/s**

difference in speed = 29 -25 = 4 m/s answer

let me know in a comment if there is any problem or doubts

difference in speed =

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