Answer
Please see the work below.
Work Step by Step
We know that the acceleration of the first motion is given as:
$a_1=\frac{0.96}{0.42}=2.2857\frac{m}{s^2}$
The frictional force is
$f=\mu_1 mg=ma_1$
This can be rearranged as:
$\mu_1=\frac{a_1}{g}$
We plug in the known values to obtain:
$\mu_1=\frac{2.2857}{9.8}=0.23$
The acceleration of the second motion is given as:
$a_1=\frac{0.96}{0.33}=2.909\frac{m}{s^2}$
The frictional force is
$f=\mu_2 mg=ma_2$
This can be rearranged as:
$\mu_2=\frac{a_2}{g}$
We plug in the known values to obtain:
$\mu_2=\frac{2.909}{9.8}=0.30$
Thus, the frictional coefficient is between 0.23 and 0.30.
