Answer
(a) $24Kg$
(b) $0.27$
Work Step by Step
(a) We can find the mass of the box as follows:
$\Sigma F_x=-\mu_kmg +81N=m(0.75m/s^2)$...... eq(1)
$\Sigma F_x=-\mu_kmg +75N=m(0.50m/s^2)$...... eq(2)
Subtracting eq(2) from eq(1), we obtain:
$0+(81-75)N=m(0.75-0.50)m/s^2$
$\implies m=\frac{6N}{0.26m/s^2}$
$m=24Kg$
(b) We can find the required coefficient of kinetic friction as follows:
$\mu_k=\frac{75-m(0.50)}{mg}$
We plug in the known values to obtain:
$\implies \mu_k=\frac{75-(24)(0.50)}{(24)(9.81)}$
$\mu_k=0.27$
