Respuesta :
Let's consider all the forces acting on the direction parallel to the surface of the slide. We have in total two forces in this direction:
1) the component of the weight parallel to the surface, which is equal to
[tex]W_{//}= mg \sin \alpha[/tex]
where [tex]\alpha =27.0 ^{\circ}[/tex], directed toward the bottom of the slide;
2) The frictional force, which is given by
[tex]F_f = \mu m g \cos \alpha[/tex]
where [tex]\mu [/tex] is the coefficient of kinetic friction that we want to find, and this force is directed towards the top of the slide, so in the opposite direction of the previous force.
For Newton's second law, the algebraic sum of the forces acting in this direction must be equal to the product of the mass and the acceleration of the child:
[tex]\sum F = ma[/tex]
So, if we explicite the forces, this becomes
[tex]mg \sin \alpha - \mu mg \cos \alpha = ma[/tex]
from which we can isolate [tex]\mu[/tex], and by using the information about the acceleration ([tex]a=1.12 m/s^2[/tex]), we find:
[tex]\mu = \frac{g \sin \alpha -a}{g \cos \alpha}= \frac{(9.81 m/s^2)(\sin 27^{\circ})-1.12 m/s^2}{(9.81 m/s^2)(\cos 27^{\circ})} =0.38 [/tex]
1) the component of the weight parallel to the surface, which is equal to
[tex]W_{//}= mg \sin \alpha[/tex]
where [tex]\alpha =27.0 ^{\circ}[/tex], directed toward the bottom of the slide;
2) The frictional force, which is given by
[tex]F_f = \mu m g \cos \alpha[/tex]
where [tex]\mu [/tex] is the coefficient of kinetic friction that we want to find, and this force is directed towards the top of the slide, so in the opposite direction of the previous force.
For Newton's second law, the algebraic sum of the forces acting in this direction must be equal to the product of the mass and the acceleration of the child:
[tex]\sum F = ma[/tex]
So, if we explicite the forces, this becomes
[tex]mg \sin \alpha - \mu mg \cos \alpha = ma[/tex]
from which we can isolate [tex]\mu[/tex], and by using the information about the acceleration ([tex]a=1.12 m/s^2[/tex]), we find:
[tex]\mu = \frac{g \sin \alpha -a}{g \cos \alpha}= \frac{(9.81 m/s^2)(\sin 27^{\circ})-1.12 m/s^2}{(9.81 m/s^2)(\cos 27^{\circ})} =0.38 [/tex]
The friction force is the product of the coefficient of friction and normal reaction. The coefficient of kinetic friction between the child and the slide will be 0.38.
What is the friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
By resolving all the forces we get
Force in the y-direction
[tex]\rm W = mgsin\alpha[/tex]
Force in the x-direction
[tex]\rm R = mgcos\alpha[/tex]
[tex]f= \mu mgcos\alpha[/tex]
According to newtons second law of motion
∑F = ma
∑F = mgsinα-μmgcosα
mgsinα-μmgcosα=ma
a=1.12 m/sec²
[tex]\rm{ \mu = \frac{gsin\alpha -a}{gcos\alpha } }[/tex]
[tex]\rm{ \mu = \frac{9.81\timessin27^0 -a}{9.81\times cos27^0 } }\\\\\mu = 0.38[/tex]
Hence the coefficient of kinetic friction between the child and the slide will be 0.38.
To know more about friction force refer to the link;
https://brainly.com/question/1714663
