Question Video: Finding the Distance Covered by a Body on a Rough Inclined Plane until It Comes to Rest Mathematics

A body of mass 74 kilograms was projected at 8.5 m/s along the line of greatest slope up a plane inclined at 30° to the horizontal. Given that the resistance of the plane to its motion was 7.4 N, find the distance the body traveled until it came to rest. Take 𝑔 = 9.8 m/s².

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Video Transcript

A body of mass 74 kilograms was projected at 8.5 meters per second along the line of greatest slope up a plane inclined at 30 degrees to the horizontal. Given that the resistance of the plane to its motion was 7.4 newtons, find the distance the body traveled until it came to rest. Take 𝑔 to equal 9.8 meters per second squared.

Alright, so let’s say that this is our slope with its inclination angle of 30 degrees. And on this slope, we have a body being projected along the slope line at 8.5 meters per second. As the body moves up the slope, it encounters a frictional force we can call 𝐹 with a given magnitude of 7.4 newtons. We’re told further that the body has a mass we’ll call 𝑚 of 74 kilograms. And knowing that the body eventually comes to a rest, we want to calculate the distance along the slope, we’ll call it 𝑑, that it takes to do this.

As we get started, let’s clear a bit of space on screen and recall that Newton’s second law of motion tells us that the net force acting on a body is equal to its mass times its acceleration. Considering an up-close view of our body as it moves up the incline, let’s draw a free-body diagram showing all the forces acting on it. We know that the body experiences a weight force, its mass times the acceleration due to gravity, a normal or reaction force perpendicular to the surface of the plane. And because of the body is moving uphill, it experiences a frictional force downhill. These are the forces that act on our body as it rises up the incline.

And let’s say that, directionally, the positive 𝑥-direction points down the incline and the positive 𝑦-direction points perpendicularly away from it. Applying Newton’s second law to this scenario, we can say that the sum of forces in the 𝑥-direction equals our body’s mass times its acceleration in this direction. These forces include the frictional force 𝐹, as well as a component of the weight force that acts in the 𝑥-direction.

To solve for this component, we need to realize that this angle in our right triangle here is identical to the 30-degree angle of inclination of our plane. Because that’s true, the component of the weight force we’re interested in is 𝑚 times 𝑔 times the sin of 30 degrees. The sin of 30 degrees is one-half. So, when we write out all the forces acting on our body in the 𝑥-direction, we have 𝐹 plus 𝑚 times 𝑔 divided by two. The second law tells us that this sum equals our body’s mass times its acceleration in-in this case, the 𝑥-direction. Dividing both sides of this equation by 𝑚, we find that factor cancelling on the right. And we find that 𝑎 sub 𝑥 is equal to 𝐹 plus 𝑚 times 𝑔 over two all divided by the mass 𝑚.

Since we’re given the force 𝐹, the mass 𝑚, and we also know that the constant 𝑔 is equal to 9.8 meters per second squared, we can substitute those values into this equation to solve for 𝑎 sub 𝑥. When we enter this expression on our calculator, we find a result of exactly five with units of meters per second squared. This means that as our body moves up the incline, it decelerates at 5 meters per second squared. And this will help us solve for the distance 𝑑 it takes to come to a rest.

Because our body experiences a constant acceleration, that means its motion can be described by what are called the equations of motion. These are sometimes also called the kinematic equations or SUVAT equations. And in our case, we’ll look at a specific equation of motion that tells us that the final velocity of a body squared equals its original velocity squared plus two times its acceleration multiplied by its displacement. Rearranging this equation to solve for 𝑑, it’s equal to 𝑣 sub 𝐹 squared minus 𝑣 sub zero squared all over two 𝑎.

And as we consider our scenario with respect to these variables, we know that 𝑣 sub 𝐹 is zero because our body ends up at rest. We’re told the value of what’s called here 𝑣 sub zero. That’s 8.5 meters per second. And we’ve just recently solved for the acceleration 𝑎. If we plug in these values and leave out units for now, we get this expression. And notice that it’s a negative value. The reason for this is that, technically, we’re solving for displacement, which, according to our sign convention, where positive values are down the incline is actually truly a negative value.

What we actually want to solve for, though, is the distance that our body travels before it comes to rest. To calculate that in this case, we just need to take the absolute value of this fraction. Entering this expression on our calculator, we get a result of 7.225. The units for this distance are meters. So, we can say that this body moves 7.225 meters up this incline before it comes to a stop.

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