# 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/s². Determine the coefficient of friction of the plane. Take 𝑔 = 9.8 m/s².

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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 𝑔 equal to 9.8 meters per second squared.

We will begin by drawing a free-body diagram to represent the scenario in this question. We’re told that the body has a mass of 21 kilograms. There will therefore be a downward force of 21𝑔, where 𝑔 is the acceleration due to gravity equal to 9.8 meters per second squared. Due to Newton’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 𝐅 𝑟 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 𝜇 is equal to the frictional force 𝐅 𝑟 divided by the normal reaction force 𝑅. We will therefore begin by calculating these values by resolving horizontally and vertically.

Newton’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 𝐅 𝑟. 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 𝐅 𝑟 is equal to 41.16 The frictional force is 41.16 newtons.

We can calculate the value of the normal reaction force 𝑅 by resolving vertically. If we take the positive direction to be vertically upwards, the sum of our forces is 𝑅 minus 21𝑔. As the body is not moving in this direction, this is equal to zero. Adding 21𝑔 to both sides of our equation, we have 𝑅 is equal to 21𝑔. 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 𝜇. 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.