# Question Video: Finding the Coefficient of Friction of a Body on a Rough Horizontal Plane Moving with Uniform Acceleration under a Horizontal Force Mathematics

A body of mass 21 kg was placed on a rough horizontal plane. A horizontal force of magnitude 144.06 N acted on the body, causing a uniform acceleration of 4.9 m/sΒ². Determine the coefficient of friction of the plane. Take π = 9.8 m/sΒ².

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### Video Transcript

A body of mass 21 kilograms was placed on a rough horizontal plane. A horizontal force of magnitude 144.06 newtons acted on the body, causing a uniform acceleration of 4.9 meters per second squared. Determine the coefficient of friction of the plane. Take π equal to 9.8 meters per second squared.

We will begin by drawing a free-body diagram to represent the scenario in this question. Weβre told that the body has a mass of 21 kilograms. There will therefore be a downward force of 21π, where π is the acceleration due to gravity equal to 9.8 meters per second squared. Due to Newtonβs third law, we will have a normal reaction force acting in the opposite direction to this. This will act vertically upwards. We are told that a horizontal force of magnitude 144.06 newtons acts on the body.

As the horizontal plane is rough, there will be a frictional force π π acting in the opposite direction to the motion. Finally, we are told that the body moves with a uniform acceleration of 4.9 meters per second squared. We are asked to calculate the coefficient of friction of the plane. This coefficient of friction π is equal to the frictional force π π divided by the normal reaction force π. We will therefore begin by calculating these values by resolving horizontally and vertically.

Newtonβs second law states that the sum of the forces is equal to mass multiplied by acceleration. In the horizontal direction, we have the 144.06-newton force and the frictional force. If we take the direction of motion to be positive, we have 144.06 minus π π. This is equal to the mass of 21 kilograms multiplied by the acceleration of 4.9 meters per second squared. The right-hand side of our equation simplifies to 102.9. We can then add the frictional force and subtract 102.9 from both sides of our equation such that π π is equal to 41.16 The frictional force is 41.16 newtons.

We can calculate the value of the normal reaction force π by resolving vertically. If we take the positive direction to be vertically upwards, the sum of our forces is π minus 21π. As the body is not moving in this direction, this is equal to zero. Adding 21π to both sides of our equation, we have π is equal to 21π. Multiplying 21 by 9.8 gives us a normal reaction force of 205.8 newtons.

We can now use both of these values to calculate the coefficient of friction π. It is equal to 41.16 divided by 205.8, which simplifies to one-fifth or 0.2. The coefficient of friction of the plane is one-fifth.