Answer
(a) $a = 3.62~m/s^2$
$F_f = 4.83~N$
$\mu_s = 0.313$
(b) $a = 3.62~m/s^2$
$F_f = 9.66~N$
$\mu_s = 0.313$
Work Step by Step
(a) If the ball rolls without slipping, then $\alpha = \frac{a}{R}$. We can use a torque equation to find an expression for the force of static frction $F_f$.
$\tau = I\alpha$
$F_f~R = (\frac{2}{3}mR^2)(\frac{a}{R})$
$F_f = \frac{2ma}{3}$
We can use this expression in the force equation for the ball as it moves down the slope.
$\sum F = ma$
$mg~sin(\theta) - F_f = ma$
$mg~sin(\theta) - \frac{2ma}{3} = ma$
$\frac{5a}{3} = g~sin(\theta)$
$a = \frac{3g~sin(\theta)}{5}$
$a = \frac{(3)(9.80~m/s^2)~sin(38.0^{\circ})}{5}$
$a = 3.62~m/s^2$
We can use the acceleration to find the force of static friction.
$F_f = \frac{2ma}{3}$
$F_f = \frac{(2)(2.00~kg)(3.62~m/s^2)}{3}$
$F_f = 4.83~N$
We can use the force of static friction to find the minimum coefficient of static friction.
$F_f = 4.83~N$
$mg~cos(\theta)~\mu_s = 4.83~N$
$\mu_s = \frac{4.83~N}{mg~cos(\theta)}$
$\mu_s = \frac{4.83~N}{(2.00~kg)(9.80~m/s^2)~cos(38.0^{\circ})}$
$\mu_s = 0.313$
(b) Since the acceleration does not depend on mass, the acceleration would not change.
$a = 3.62~m/s^2$
Since $F_f = \frac{2ma}{3}$, if the mass doubles, then the force of static friction doubles.
$F_f = 2\times (4.83~N) = 9.66~N$
Since $\mu_s$ does not depend on the mass, the minimum value of $\mu_s$ is still $\mu_s = 0.313$.
