# Question Video: Finding the Time Taken by a Body Projected on a Rough Inclined Plane to Come to Rest Mathematics

A body of mass 30 kg was projected at 12 m/s along the line of greatest slope of a plane inclined at 30° to the horizontal. Given that the resistance of the plane to the movement of the body was 3 N, how long did it take for the body to come to rest? Consider the acceleration due to gravity to be 9.8 m/s².

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### Video Transcript

A body of mass 30 kilograms was projected at 12 meters per second along the line of greatest slope of a plane inclined at 30 degrees to the horizontal. Given that the resistance of the plane to the movement of the body was three newtons, how long did it take for the body to come to rest? Consider the acceleration due to gravity to be 9.8 meters per second squared.

Okay, so let’s say that this is our plane inclined at an angle of 30 degrees and that this is our body, which we’re told is projected up the plane in this direction. Its initial speed, what we’ll call 𝑣 sub 𝑖, is 12 meters per second. And because there’s resistance to this body as it moves up the plane, it will slow down and then eventually come to a stop. In this exercise, we want to figure out how long this takes. Let’s let 𝑡 represent this amount of time.

And to get started solving for it, we can recognize that since our body starts out at 12 meters per second and ends up at rest, it undergoes an acceleration. We can begin solving for the magnitude of this acceleration by recalling Newton’s second law of motion. This law tells us that the net force acting on a body is equal to that body’s mass times its acceleration. Let’s draw an up-close view of our body as it moves up this plane. We know that as it does so, it’s acted on by a weight force, 𝑚 times 𝑔, a normal or a reaction force, we’ll call 𝑅, and also a frictional force.

Since in this snapshot in time, we imagine our body is still moving up the incline, we know that the frictional force will point the other way down the slope. Building on this free body diagram, we can decide that the positive 𝑥-direction points down the slope and that the positive 𝑦-direction points perpendicularly away from it. What we’re going to do now is apply Newton’s second law of motion to the forces on our body in the 𝑥-direction. We’ll make some space onscreen to do that. But before we do, let’s record the fact that the mass of our body, we’ll call it 𝑚, is 30 kilograms and that the frictional force opposing the body’s motion up the slope is three newtons.

So now, using our diagram, let’s consider the forces on our body in the 𝑥-direction. By our sign convention that the positive 𝑥-direction is down the slope, we can say that our frictional force has a positive value. And then breaking the weight force on our body up into its 𝑥- and 𝑦-components, we can see that this section here represents that 𝑥-component.

Looking at this right triangle made by the components of the weight force, it turns out that this angle right here is equal to this interior angle of our original slope triangle. That is, it also is 30 degrees. This tells us that the component of the weight force that acts in the 𝑥-direction is 𝑚 times 𝑔 times the sin of 30 degrees. And let’s recall that the sin of 30 degrees is one-half. This means we can add 𝑚 times 𝑔 divided by two to our frictional force 𝐹. And since these are the only forces acting on our body in the 𝑥-direction, their sum is equal to the body’s mass times its acceleration in this dimension.

If we then divide both sides of this equation by our body’s mass, that factor cancels on the right and we wind up with this expression for our body’s acceleration along the incline. In our problem statement, we were told the mass of our body as well as the frictional force it experiences. And we also know that the acceleration due to gravity is 9.8 meters per second squared. Plugging these values without their units into our expression, the acceleration of our body along the slope comes out to exactly five, where the units are meters per second squared.

So coming back to our original sketch, our body, which started out with a speed of 12 meters per second, decelerates at five meters per second squared until it comes to rest. The question is, how long does this process take? Because our body undergoes a constant acceleration, that means that it can be described by what are called the equations of motion. These equations describe the motion of any body under constant acceleration.

Though there are four equations of motion, we’ll focus on this one that tells us that the final velocity of an object is equal to its initial velocity plus its acceleration times the elapsed time. In our situation, since our body ends up at rest, we can say that 𝑣 sub 𝑓 is equal to zero and then 𝑣 sub 𝑖, its initial velocity, is negative because we’ve defined down the slope as the positive direction. To this, we add our acceleration 𝑎 times the time we want to solve for 𝑡.

Our equation of motion tells us that this left- and right-hand side are equal. And so if we add 12 to both sides, we get five 𝑡 is equal to 12. And this implies that 𝑡 itself is equal to twelve-fifths. Written as a decimal, this is 2.4. And we know it has units of seconds. So it takes 2.4 seconds for this body projected up this incline at 12 meters per second to come to rest.

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