A block of mass 2 kg is placed on inclined rough surface AC (as shown in figure) of coefficient of friction $\mathrm{\mu }$. If g = 10 ms^-2, the net force (in N) on the block will be :

Consider a block of mass 2 kg placed on an inclined rough surface AC. The coefficient of friction for the surface is given as μ = 1/3. We need to calculate the net force acting on the block.

Since the block of mass 2 kg is placed on an inclined rough surface AC, two main forces act on it: the component of gravitational force along the incline and the frictional force opposing the motion. The gravitational force component is calculated as:

F = mg sin 30°

The frictional force opposing the motion is calculated as:

Friction Force = μmg cos 30°

Since the block of mass 2 kg is placed on an inclined rough surface AC, the net force acting on the block can be calculated by subtracting the frictional force from the gravitational component:

Net Force (F) = mg sin 30° - μmg cos 30°

Substituting the given values:

F = (2)(10)(1/2) - (1/3)(2)(10)(√3/2)

Simplifying the calculations:

F = 10 - (10√3 / 3)

Since √3 ≈ 1.732, the calculation becomes:

F = 10 - (10 × 1.732 / 3)

F = 10 - 5.77

F ≈ 0

Therefore, when a block of mass 2 kg is placed on an inclined rough surface AC, the net force acting on the block becomes zero. This implies that the block will remain in equilibrium without any net movement.

To conclude, a block of mass 2 kg placed on an inclined rough surface AC experiences a net force of zero due to the balance between the gravitational component and the frictional force.