A block of mass 1 kg slides down on a rough inclined plane of inclination 60° starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is 1 m, then work done against friction is (Take g = 9.8 m/s²

The correct answer is 2.45 J.

When a block slides down a rough inclined plane, friction acts upward along the surface of the plane (opposing the motion). The work done against friction is the energy lost due to the frictional force over the length of the plane.

Step 1: Identify the given values

Mass of block: $m=1\text{kg}$
Angle of inclination: $\theta =\mathrm{60°}$
Coefficient of kinetic friction: ${\mu }_k=0.5$
Length of inclined plane: $L=1\text{m}$
Acceleration due to gravity: $g=9.8\text{m/s²}$

Step 2: Find the Normal Force

On an inclined plane, the weight of the block acts vertically downward. This weight has two components — one along the incline (mg sin θ) and one perpendicular to the incline (mg cos θ). The surface can only push back perpendicular to itself, so the normal force $N$ balances the perpendicular component:

$N=mgcos\theta$

$N=1\times 9.8\times cos\mathrm{60°}$

Since $cos\mathrm{60°}=0.5$:

$N=9.8\times 0.5=4.9\text{N}$

Step 3: Find the Kinetic Friction Force

The kinetic friction force is given by:

$f_k={\mu }_k\times N$

$f_k=0.5\times 4.9=2.45\text{N}$

Step 4: Calculate Work Done Against Friction

Work done against friction is the product of the frictional force and the displacement along the surface of the incline. Since friction opposes the motion, the work done against it is positive:

$W_{\mathrm{friction}}=f_k\times L$

$W_{\mathrm{friction}}=2.45\times 1=2.45\text{J}$

Summary of Key Concept: The normal force on an inclined plane is $mgcos\theta$, not $mg$. Since cos 60° = 0.5, the normal force is halved compared to a flat surface, which reduces the friction force and consequently the work done against friction.

Therefore, the work done against friction is 2.45 J.