We are tasked with proving that the coefficient of kinetic friction $\mu_k = \tan \phi$ for an object sliding down an inclined plane at constant speed. Here's the step-by-step approach:

1. Forces Acting on the Object:

   • Gravity causes the object to accelerate down the incline, with a component along the incline of $M g \sin \phi$, where $M$ is the mass of the object, $g$ is the acceleration due to gravity, and $\phi$ is the incline angle.
   • The normal force perpendicular to the surface is $N = M g \cos \phi$.
   • Kinetic friction acts opposite to the motion, with a magnitude $f_k = \mu_k N = \mu_k M g \cos \phi$.

2. Constant Speed Condition: Since the object is moving at a constant speed, the net force along the incline must be zero. This gives us the equation: $M g \sin \phi - f_k = 0$ Substituting $f_k = \mu_k M g \cos \phi$, we get: $M g \sin \phi = \mu_k M g \cos \phi$

3. Solving for $\mu_k$: Canceling $M g$ from both sides (assuming $M \neq 0$ and $g \neq 0$) gives: $\sin \phi = \mu_k \cos \phi$ Dividing both sides by $\cos \phi$ gives: $\mu_k = \frac{\sin \phi}{\cos \phi} = \tan \phi$

Thus, we have shown that the coefficient of kinetic friction $\mu_k = \tan \phi$ when the object slides down the incline at a constant speed.

Would you like more details, or do you have any further questions?

Here are some related questions to deepen your understanding:

1. How would the expression change if the object was accelerating down the plane?
2. What happens to the coefficient of kinetic friction if the angle $\phi$ increases?
3. How does this relationship $\mu_k = \tan \phi$ apply to everyday situations, like driving on a slope?
4. What is the difference between static and kinetic friction, and how does it affect the motion?
5. Could the coefficient of friction ever exceed 1? If so, under what circumstances?

Tip: Understanding free-body diagrams can greatly simplify analyzing forces in mechanics problems!