Respuesta :
Explanation:
The given data is as follows.
a = 0.82 [tex]m/s^{2}[/tex], Angle ([tex]\theta[/tex]) = [tex]38^{o}[/tex]
So, in x-direction the force will act on the box. Hence, formula for normal force and x-component of the gravitational force is as follows.
[tex]F_{n} = mg Cos (\theta)[/tex] ............. (1)
In y-direction,
[tex]mg sin (\theta) - f = ma[/tex] .............. (2)
and, f = [tex]\mu_{k} F_{n}[/tex] ........... (3)
Now, we will use equations (1), and (3) in equation (2) as follows.
[tex]mg sin (\theta) - (\mu_{k} mg cos (\theta)[/tex] = ma
Then,
[tex]\mu_{k} = \frac{a - g sin (\theta)}{g cos (\theta)}[/tex]
= [tex]\frac{9.8 sin (38) - 0.82}{9.8 cos (38)}[/tex]
= [tex]\frac{2.9008 - 0.82}{9.359}[/tex]
= 0.22
Thus, we can conclude that the coefficient of kinetic friction between the box and the ramp is 0.22.
