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 =