Answer
$T = \frac{u_k}{\sqrt{1+\mu_k^2}}$
Work Step by Step
In order to find the expression for the minimum tension needed, we take the derivative of the equation of the previous problem, which is:
$\frac{T}{mg}=\frac{\mu_k}{cos\theta+\mu_ksin\theta}$
Taking the derivative gives the answer:
$T = \frac{u_k}{\sqrt{1+\mu_k^2}}$
