Answer:
[tex]\mu = 0.31[/tex]
Explanation:
Given data:
mass = 2.00 kg
slope angle = 38.0
From figure
balancing force
[tex]mgsin\theta - f = ma[/tex] .....1
Balancing torque
[tex]F_R = \frac{2}{3} mR^2 \alpha[/tex] ......2
for pure rolling
[tex]\alpha = \frac{a}{R}[/tex]
[tex]F = \frac{2}{3} ma[/tex]
from 1 and 2nd equation
[tex]mgsin\theta - \frac{2}{3}ma = ma[/tex]
[tex]mgsin\theta = \frac{5}{3} ma[/tex]
[tex]a = \frac{3}{5} g sin\theta[/tex]
[tex] = \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2[/tex]
[tex]F =\mu N[/tex]
[tex]= \frac{2}{3} ma [/tex]
[tex]= \frac{2}{3} 2\times 3.62 = 4.83 N[/tex]
N =normal force [tex]= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N[/tex]
[tex]\mu \times 15.44 = 4.83[/tex]
solving for coefficent of friction we get
[tex]\mu = 0.31[/tex]
Based on the calculations, the acceleration of this hollow, spherical shell is equal to 3.62 [tex]m/s^2[/tex].
Given the following data:
The torque for a hollow, spherical shell is given by this formula:
[tex]F=\frac{2}{3} ma[/tex]
From Newton's Second Law of Motion, the forces acting on the shell is given by:
[tex]mgsin\theta -F=ma\\\\mgsin\theta -\frac{2}{3} ma=ma\\\\mgsin\theta=ma+\frac{2}{3} ma\\\\mgsin\theta=\frac{5}{3} ma\\\\gsin\theta=\frac{5}{3} a\\\\a=\frac{3gsin\theta}{5} \\\\a=\frac{3 \times 9.8 \times sin38.0}{5} \\\\a = \frac{18.10}{5}[/tex]
Acceleration, a = 3.62 [tex]m/s^2[/tex]
Mathematically, the friction force acting on this shell is equal to the torque:
[tex]F=\frac{2}{3} ma\\\\F=\frac{2}{3} \times 2.00 \times 3.62[/tex]
F = 4.83 Newton.
First of all, we would determine the normal force:
[tex]N=mgcos\theta\\\\N= 2 \times 9.8 \times cos 38.0[/tex]
N = 15.45 Newton.
For the coefficient of friction:
[tex]\mu = \frac{F}{N} \\\\\mu = \frac{4.83}{15.45}\\\\\mu=0.31[/tex]
Read more on force here: brainly.com/question/1121817
