A child slides down a slide with a 20° incline, and at the bottom her speed is precisely one-fourth what it?

Question

A child slides down a slide with a 20° incline, and at the bottom her speed is precisely one-fourth what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child

Answer 1

Poor child.

Ft=Fd-f
Ft= mgsin(20) - umgcos(20)
Ft=mg(sin(20) - u cos(20))

a=Ft/m= g(sin(20) - u cos(20))


V=at
t=sqrt(2S/a)

At u=0 no friction
Ft=mg(sin(20)
a'=Ft/m
a' = g sin(20)
V'=a't'
t'=sqrt(2S/a')

4V=V'
4at=a't'
4asqrt(2S/a)=a' sqrt(2S/a')
4sqrt(a)=sqrt(a')
16a=a'
16[ g(sin(20) - u cos(20))]=g sin(20)
u=[16sin(20) - sin(20) ]/(16 cos(20))
u= 15/16 (tan(20)
u=.341

Answer 2

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