Block Acceleration with Friction

This problem involves analyzing the motion of a block on a surface subject to both an applied force and friction. We need to determine the block's linear acceleration by considering all the forces acting on it and applying Newton's second law of motion.

Given Parameters

We are given the following information:

• Mass of the block, \(m = 10 \text{ kg}\)
• Applied external force, \(F_{\text{applied}} = 20 \text{ N}\)
• Coefficient of friction between the block and the ground, \(\mu = 0.1\)
• Acceleration due to gravity, \(g = 10 \text{ m/s}^2\)

Forces Acting on the Block

Several forces act on the block:

• Applied Force: A horizontal force of 20 N is applied, pushing the block.
• Gravitational Force (Weight): The Earth pulls the block downwards with a force \(W = mg\).
• Normal Force: The ground pushes upwards on the block, perpendicular to the surface, counteracting the weight. On a horizontal surface, the normal force \(N\) is equal in magnitude to the weight \(W\).
• Friction Force: Since the block is sliding, there is kinetic friction opposing the motion. The friction force \(f_k\) acts horizontally, opposite to the direction of the applied force. Its magnitude is given by \(f_k = \mu N\).

Calculating Normal Force

The normal force \(N\) is equal to the weight of the block because the surface is horizontal and there are no other vertical forces.

The weight \(W\) is calculated as:

\(W = mg\)

Substituting the given values:

\(W = (10 \text{ kg}) \times (10 \text{ m/s}^2)\)

\(W = 100 \text{ N}\)

So, the normal force is:

\(N = 100 \text{ N}\)

Calculating Friction Force

The kinetic friction force \(f_k\) is calculated using the coefficient of friction and the normal force:

\(f_k = \mu N\)

Substituting the values \(\mu = 0.1\) and \(N = 100 \text{ N}\):

\(f_k = 0.1 \times 100 \text{ N}\)

\(f_k = 10 \text{ N}\)

This friction force opposes the applied force.

Applying Newton's Second Law

Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration (\(F_{\text{net}} = ma\)). In the horizontal direction, the net force is the difference between the applied force and the friction force.

The net horizontal force \(F_{\text{net}}\) is:

\(F_{\text{net}} = F_{\text{applied}} - f_k\)

Substituting the values \(F_{\text{applied}} = 20 \text{ N}\) and \(f_k = 10 \text{ N}\):

\(F_{\text{net}} = 20 \text{ N} - 10 \text{ N}\)

\(F_{\text{net}} = 10 \text{ N}\)

Determining Linear Acceleration

Now, we can find the linear acceleration \(a\) using Newton's second law:

\(F_{\text{net}} = ma\)

Rearranging the formula to solve for acceleration:

\(a = \frac{F_{\text{net}}}{m}\)

Substituting the net force \(F_{\text{net}} = 10 \text{ N}\) and mass \(m = 10 \text{ kg}\):

\(a = \frac{10 \text{ N}}{10 \text{ kg}}\)

\(a = 1 \text{ m/s}^2\)

Final Answer

The linear acceleration of the block is 1 m/s².