This question involves the concepts of the law of conservation of energy and frictional energy.
a) The speed of the child when he reaches the bottom of the slide is "6 m/s".
b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".
a)
According to the law of conservation of energy in this situation:
Loss in Potential Energy = Gain in Kinetic Energy
[tex]mgh = \frac{1}{2}mv^2\\\\2gh=v^2\\v=\sqrt{2gh}\\\\[/tex]
where,
v = velocity = ?
g = acceleration due to gravity = 9.81 m/sΒ²
h = height lost = 6 ft = 1.83 m
Therefore,
[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]
v = 6 m/s
b)
Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:
Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy
[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]
where,
l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]
[tex]\mu_k[/tex] = coefficient of kinetic friction = ?
Therefore,
[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
