We need first apply Kinetic energy law to find the distance done for the particles,
After that we can through the Work Formula find the Force, and with this force reach the coefficient of Kinetic friction.
A) We know that Kinetic Energy equation is given by,
[tex]\Delta KE = \frac{1}{2}m(v_i^2-v_f^2)[/tex]
[tex]\Delta KE = \frac{1}{2} (0.4) ((8)^2-(6)^2)[/tex]
[tex]\Delta KE = 5.6J[/tex]
B) We know that the Kinetic Frictional Force is given by,
[tex]F_f = \mu_k N[/tex]
Where [tex]\mu_k[/tex] is the coefficient of Kinetic friction and N the Normal (mg)
The work done by friction is equal to the Kinetic Energy, and the distance where this Work is applied is
[tex]d= 2\pi(1.5)=9.42[/tex]m
[tex]\Delta KE = W[/tex]
and
[tex]W=F_f*d[/tex]
Re-arrange for [tex]F_f[/tex],
[tex]F_f = \frac{W}{d} = \frac{5.6}{9.42} = 0.594N[/tex]
Solving in the equation of Kinetic Frictional Force to find the coefficient we have,
[tex]0.594N = \mu_K (0.4)(9.8)[/tex]
[tex]\mu_K = 0.102[/tex]
c) The number of total revolutions is given by the initial KE, that is
[tex]KE_i = \frac{1}{2}v_i^2 =\frac{1}{2}(0.4)*64 = 12.8[/tex]
[tex]nF_fd=12.8J[/tex], then
[tex]n=\frac{12.8}{0.594*9.42}=2.29Rev[/tex]
