Question

A block of mass m slides down with uniform speed on an inclined plane having inclination \( \theta \). If the coefficient of friction between the inclined plane and the block is \( \mu \), then the contact force between them is:

• A. \( mg \sin \theta \)
• B. \( mg \)
• C. \( \sqrt{(mg \sin \theta)^2 + (\mu mg \cos \theta)^2} \)
• D. \( mg \cos \theta \sqrt{1 + \mu^2} \)

Hint:

For uniform motion, the net force must be zero. The contact force is the resultant of the normal and frictional forces.

Answer 1

The Correct Option is D

When the block slides down with uniform speed, the forces acting along the plane and perpendicular to the plane must balance.
The contact force is the resultant of the normal force and the frictional force.
The frictional force is given by: \[ f = \mu mg \cos \theta \] The component of gravitational force along the plane is: \[ F_{\text{gravity}} = mg \sin \theta \] Since the block moves with uniform speed, the net force along the plane must be zero, so the contact force is the resultant of these two forces: \[ F_{\text{contact}} = \sqrt{(mg \sin \theta)^2 + (\mu mg \cos \theta)^2} \]
Thus, the correct answer is (d).