A brick of mass 2 kg begins to slide down on a plane inclined at an angle of 45° with the horizontal. The force of friction will be

The correct answer is 19.6 sin 45°.

When a brick slides down an inclined plane, we need to analyze all the forces acting on it carefully. Let's build the solution step by step.

Step 1: Identify the given values

Mass of brick: m = 2 kg
Angle of inclination: θ = 45°
Acceleration due to gravity: g = 9.8 m/s²
Weight of brick: W = mg = 2 × 9.8 = 19.6 N

Step 2: Resolve the weight into components along the inclined plane

When an object rests on an inclined surface, its weight (acting vertically downward) is resolved into two perpendicular components:

● Along the incline (down the slope):
W_{\parallel }=mgsin\theta =19.6sin\mathrm{45°}
This component drives the brick down the slope.

● Perpendicular to the incline (into the surface):
W_{\perp }=mgcos\theta =19.6cos\mathrm{45°}
This component determines the Normal force: N = mg cos 45°.

Step 3: Apply the condition "begins to slide"

The phrase "begins to slide" is the critical clue. It means the brick is on the verge of motion — the system is at the threshold between static equilibrium and sliding. At this exact point, the static friction force reaches its maximum value and is just enough to balance the component of weight pulling the brick down the incline.

Applying equilibrium along the incline at the moment of impending motion:

\text{Friction Force (f)}=\text{Component of weight along incline}

f=mgsin\theta

f=19.6\times sin\mathrm{45°}

f=19.6sin\mathrm{45°}\approx 13.86\text{N}

Step 4: Why not the other options?

● 19.6 cos 45° is the Normal force (N), not friction.
● 9.8 sin 45° would correspond to a mass of 1 kg, but m = 2 kg here.
● 9.8 cos 45° is neither friction nor normal force for a 2 kg brick.

Summary: The force of friction at the moment the brick begins to slide is equal to the driving component of gravity along the slope, which is f = mg sin 45° = 19.6 sin 45° ≈ 13.86 N, directed up the incline, opposing the tendency of motion.