Question

A block has been placed on a inclined plane with the slope angle $\theta$, block slides down the plane at constant speed. The coefficient of kinetic friction is equal to

• A. $\sin \theta$
• B. $\cos \theta$
• C. $g$
• D. \(\tan \theta\)

Answer 1

The Correct Option is D

To determine the coefficient of kinetic friction for a block sliding down an inclined plane at constant speed, we can use the principles of basic mechanics. When the block slides at constant speed, it means that the net force acting along the plane is zero.

1. Consider the forces acting on the block:

   • The gravitational force acting down the slope: mg \sin \theta, where m is the mass of the block and g is the acceleration due to gravity.
   • The frictional force acting up the slope: \text{f}_{\text{k}} = \mu_{\text{k}} \cdot N, where \mu_{\text{k}} is the coefficient of kinetic friction and N is the normal force.
   • The normal force: N = mg \cos \theta.

2. Since the block is sliding at a constant speed, the forces along the inclined plane must be balanced. Therefore, the force of kinetic friction must equal the component of gravitational force down the slope:

   mg \sin \theta = \mu_{\text{k}} \cdot mg \cos \theta

3. Cancel the common factor mg from both sides:

   \sin \theta = \mu_{\text{k}} \cos \theta

4. Solve for \mu_{\text{k}}:

   \mu_{\text{k}} = \frac{\sin \theta}{\cos \theta} = \tan \theta

Thus, the coefficient of kinetic friction is \tan \theta. This matches with the given correct answer.

By understanding that balance of forces leads to the formula we derived, we can conclude that the coefficient of kinetic friction when a block slides at a constant speed on an inclined plane is indeed \tan \theta.