Discover the Coefficient of Kinetic Friction on a Slide with a 26° Incline

[ISU20CpreE]

Need explanation, please~~

A child slides down a slide with a 26° incline, and at the bottom her speed is precisely one-third what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.


I get as an answer 0.43 as the coeficient of kinetic friction.

Some ideas please post back.

Thanks.

 

[ISU20CpreE]

A child slides down a slide with a 26° incline, and at the bottom her speed is precisely one-third what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.


I get as an answer 0.43 as the coeficient of kinetic friction.

Some ideas please post back.

Thanks.

I'm so sorry for violating the rules of the site. I am a new member and I am learning from my mistakes. I solved the problem, but i don't understand some parts of the problem. This is my procedure:

The kinetic friction force will be up the slide to oppose the motion.
We choose the positive direction in the direction of the acceleration.
From the force diagram for the child, we have ?F = ma:
x-component: mg sin \theta[\tex] – Ffr = ma;<br /> y-component: FN – mg cos \theta[\tex] = 0.&lt;br /&gt; When we combine these, we get&lt;br /&gt; a = g sin \theta[\tex] – \mu[\tex]kg cos \theta[\tex] = g(sin \theta[\tex] – \mu[\tex]k cos \theta[\tex]).&amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; We can use this for the frictionless slide if we set theta[\tex]k = 0.&amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; For the motion of the child, we have&amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; v2 = v02 + 2a(x – x0) = 0 + 2ad, where d is the distance along the slide.&amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; If we form the ratio for the two slides, we get&amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; (vfriction/vnone)2 = afriction/anone = (sin \theta[\tex] – \mu[\tex]k cos \theta[\tex])/sin \theta[\tex];&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;gt; (!)2 = (sin 28° – \mu[\tex]k cos 28°)/sin 28°, which gives \mu[\tex]k = 0.40.