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 [tex]\theta[\tex] – Ffr = ma;
y-component: FN – mg cos [tex]\theta[\tex] = 0.
When we combine these, we get
a = g sin [tex]\theta[\tex] – [tex]\mu[\tex]kg cos [tex]\theta[\tex] = g(sin [tex]\theta[\tex] – [tex]\mu[\tex]k cos [tex]\theta[\tex]).
We can use this for the frictionless slide if we set [tex]theta[\tex]k = 0.
For the motion of the child, we have
v2 = v02 + 2a(x – x0) = 0 + 2ad, where d is the distance along the slide.
If we form the ratio for the two slides, we get
(vfriction/vnone)2 = afriction/anone = (sin [tex]\theta[\tex] – [tex]\mu[\tex]k cos [tex]\theta[\tex])/sin [tex]\theta[\tex];
(!)2 = (sin 28° – [tex]\mu[\tex]k cos 28°)/sin 28°, which gives [tex]\mu[\tex]k = 0.40.

 

[quicknote]

I would first like to clarify the concept of coefficient of kinetic friction. This is a measure of the amount of friction between two surfaces in contact when one of the surfaces is in motion. It is represented by the symbol "μk" and is a dimensionless quantity.

In this scenario, we are given the angle of incline of the slide and the fact that the child's speed at the bottom is one-third of what it would have been on a frictionless slide. From this information, we can use the equation μk = tan θ to calculate the coefficient of kinetic friction.

Substituting the given angle of 26° in the equation, we get μk = tan 26° = 0.4877. This means that the coefficient of kinetic friction between the slide and the child is approximately 0.4877.

However, this value may not be the same as the one calculated by the child's speed at the bottom. This could be due to factors such as the child's weight, the material of the slide, and the presence of any lubricants or debris on the surface. To obtain a more accurate value, multiple trials with different children and slides could be conducted and an average coefficient of kinetic friction could be calculated.

In conclusion, the coefficient of kinetic friction on a slide with a 26° incline can be calculated using the equation μk = tan θ. However, further experimentation and analysis may be needed for a more accurate value.