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A child slides down a slide with a 28° incline, and at the bottom her speed is precisely one half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child. here's the two equations I used: sin28*m*g=m(v2/t) since v2=v1+at the other one is with friction: sin28*m*g-u*cos28*m*g=m((1/2*v2)/t)

Respuesta :

lcmendozaf

Answer:

μ=0.13

Explanation:

If there would be no friction:

[tex]1/2*m*V_{wf}^2-m*g*L*sin\alpha =0[/tex]

Solving for the speed:

[tex]V_{wf}=\sqrt{2*g*L*sin\alpha }[/tex]

Since her speed is precisely one half what it would have been if the slide had been frictionless:

[tex]V_2=V_{wf}/2=\sqrt{2*g*L*sin\alpha } /2[/tex]

By kinetics equation:

[tex]V_2^2=V_o^2+2*a*L[/tex]   where Vo=0

Solving for a:

[tex]a=\frac{g*sin\alpha }{4}[/tex]

By a sum of forces:

[tex]\mu*N=m*a[/tex]  where [tex]N=m*g*cos\alpha[/tex]

Replacing the value for N, and a:

[tex]\mu*m*g*cos\alpha =m*\frac{g*sin\alpha }{4}[/tex]

Solving for μ:

[tex]\mu=\frac{sin\alpha }{4*cos\alpha } =0.13[/tex]