Question

A child slides down a slide with a 20 degree incline, and at the bottom her speed is percisely half what it would have been if the slide was frictionless. Derive an equation for the acceleration with friction and for acceleration without friction. Calculate the coefficient of friction between the slide and the child.

Bonus: (5pts) Must have everything correct A child slides down a slide with a 20 incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Derive an equation for the acceleration with friction and for acceleration without friction. Calculate the coefficient of friction between the slide and the child. M- ? e = 20° Vf²= 2a, Ax -No Fricsson V 2 = 4(2a,AX) - Friction a = 402

Answer 1

suppose his at the bottom in case of frictional -less = v  

take the length of the inclined plane = d

force on child = mg*sinθ

a = mg*sinθ /m =g*sinθ = 9.8*sin20 = 3.35 m/s²

accleration without friction = 3.35 m/s²

speed at bottom

v^2 = 0 +2*a*d

v^2 = 2*3.35*d ....i

case of friction  

F =mgsinθ   -μmg*cosθ

a = F/m = gsinθ   -μg*cosθ

speed at bottom

(v/2)^2 = 0 +2*(gsinθ   -μg*cosθ ) *d .... ii

from i and ii divide them

v^2 / ( v^2/4 ) = 2*3.35*d / ( 2*(gsinθ   -μg*cosθ ) *d )

4 =3.35/ gsinθ   -μg*cosθ

4 = 3.35 / ( 9.8*sin20 -μ*9.8*cos20)  

3.35 4 = 3.3518 – 9.20899μ

from here μ≈0.273026 answer

acceleration on with friction

= gsinθ   -μg*cosθ  

= 9.8*sin20- 0.273026*9.8*cos20 =

=0.837504 m/s²