A body of 5 kg weight kept on a rough inclined plane of angle 30° starts sliding with a constant velocity. Then the coefficient of friction is (assume g = 10 m/s²)

The correct answer is 1/√3.

When a body slides down a rough inclined plane with constant velocity, it means the net force on the body is zero — the body is in equilibrium. This is the key condition we will use.

Step 1: Identify the forces acting on the body

Consider a body of mass m = 5 kg on an inclined plane at angle θ = 30°. The three forces acting are:

• Weight (mg) acting vertically downward
• Normal reaction (N) acting perpendicular to the inclined surface
• Friction force (f) acting up the incline (opposing sliding motion, which is downward)

Step 2: Resolve forces along and perpendicular to the incline

Along the incline (taking down-the-slope as positive):
Component of weight along the incline = mg sin θ (down the slope)
Friction force = f (up the slope)

Perpendicular to the incline:
Normal force N = mg cos θ

Step 3: Apply the equilibrium condition (constant velocity → net force = 0)

Since the body moves with constant velocity, the net force along the incline is zero:

f=mgsin\theta

Also, friction is related to the normal force by:

f=\mu N=\mu mgcos\theta

Step 4: Equate and solve for μ

Setting the two expressions for friction equal:

\mu mgcos\theta =mgsin\theta

The mg cancels from both sides (notice the result is independent of the mass!):

\mu =\frac{sin\theta }{cos\theta }=tan\theta

Step 5: Substitute θ = 30°

\mu =tan30°=\frac{1}{\sqrt{3}}

Conclusion: The coefficient of friction is μ = 1/√3. This elegant result — μ = tan θ — is a fundamental principle: when a body slides at constant velocity down an incline, the coefficient of kinetic friction exactly equals the tangent of the angle of inclination. The mass (5 kg) and g (10 m/s²) were not needed because they cancel out.