4. A mass $m$ block slides down an inclined surface with constant speed $v$, as shown. If the coefficient of kinetic friction with the surface is $\mu_{k}$, what is the angle $\theta$ of the inclined surface?s A. $\theta=\sin ^{-1}\left(\mu_{k}\right)$ B. $\theta=\cos ^{-1}\left(\mu_{k}\right)$ C. $\theta=\tan ^{-1}\left(\mu_{k}\right)$ D. $\theta=\tan ^{-1}\left(\frac{1}{\mu_{k}}\right)$

Step 1 ：Since the block is moving at a constant speed, the net force acting on it must be zero. We can analyze the forces acting on the block in the vertical and horizontal directions.

Step 2 ：In the vertical direction, we have the gravitational force \(mg\sin\theta\) acting downward and the normal force \(N\) acting upward. In the horizontal direction, we have the gravitational force \(mg\cos\theta\) acting downward and the frictional force \(f_k = \mu_k N\) acting upward.

Step 3 ：Since the net force is zero, we can set up the following equations: \(N = mg\sin\theta\) and \(f_k = mg\cos\theta\).

Step 4 ：Substituting the expression for \(f_k\) from equation 2 into equation 1, we get: \(\mu_k N = mg\cos\theta\).

Step 5 ：Dividing both sides by \(mg\), we get: \(\mu_k = \cos\theta\).

Step 6 ：Now, we can solve for \(\theta\): \(\theta = \cos^{-1}(\mu_k)\).

Step 7 ：\(\boxed{\text{Final Answer: } B. \theta=\cos ^{-1}\left(\mu_{k}\right)}\)