A block of mass 2 kg slides down an incline plane of inclination 30°. The coefficient of friction between block and plane is 0.5. The contact force between block and plank is :

To determine the contact force acting on a block sliding down an inclined plane, we need to analyze the forces at play. In this scenario, we have a block of mass 2 kg on a 30° incline with a coefficient of friction of 0.5. Let's break down the problem step by step.

Identifying the Forces

When the block is on the incline, several forces act on it:

• Gravitational Force (Weight): This acts vertically downward and is calculated as \( W = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
• Normal Force (N): This acts perpendicular to the surface of the incline.
• Frictional Force (f): This opposes the motion of the block and is calculated using the formula \( f = \mu N \), where \( \mu \) is the coefficient of friction.

Calculating the Gravitational Force

First, we calculate the weight of the block:

W = mg = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N}

Resolving Forces Along the Incline

The weight can be resolved into two components: one parallel to the incline and one perpendicular to it.

• The component of weight parallel to the incline is given by \( W_{\parallel} = W \sin(\theta) \).
• The component of weight perpendicular to the incline is given by \( W_{\perpendicular} = W \cos(\theta) \).

Substituting the values:

W_{\parallel} = 19.62 \sin(30°) = 19.62 \times 0.5 = 9.81 \, \text{N}

W_{\perpendicular} = 19.62 \cos(30°) = 19.62 \times \frac{\sqrt{3}}{2} \approx 16.97 \, \text{N}

Calculating the Normal Force

The normal force (N) is equal to the perpendicular component of the weight since there is no vertical acceleration:

N = W_{\perpendicular} = 16.97 \, \text{N}

Determining the Frictional Force

Now, we can calculate the frictional force using the coefficient of friction:

f = \mu N = 0.5 \times 16.97 \approx 8.485 \, \text{N}

Finding the Contact Force

The contact force between the block and the incline is essentially the normal force, which we calculated earlier. Therefore, the contact force acting on the block is:

Contact Force (N) = 16.97 N

Summary

In summary, the contact force acting on the block as it slides down the incline is approximately 16.97 N. This force is crucial as it helps determine the frictional force opposing the motion of the block, allowing us to understand the dynamics of the system more thoroughly.