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Question Video: Studying the Motion of a Projected Body on a Smooth Inclined Plane Mathematics

A body of mass 9 kg is released from rest on a smooth inclined plane. It moves a distance of 25.2 m in the first 4 seconds of its motion. If the body is projected upward along a line of greatest slope on the same plane, with an initial velocity of 12.6 m/s, how far does it travel before coming to instantaneous rest? Take ๐‘” = 9.8 m/sยฒ.

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Video Transcript

A body of mass nine kilograms is released from rest on a smooth inclined plane. It moves a distance of 25.2 meters in the first four seconds of its motion. If the body is projected upward along a line of greatest slope on the same plane, with an initial velocity of 12.6 meters per second, how far does it travel before coming to instantaneous rest? Take ๐‘” to equal 9.8 meters per second squared.

Alright, so here we have a smooth inclined plane. And weโ€™re told that a body with a mass weโ€™ll call ๐‘š of nine kilograms is released from rest on the plane. Since thereโ€™s no frictional force opposing its motion, the body starts to slide downward. And weโ€™re told that after four seconds, itโ€™s moved a distance on the plane of 25.2 meters. This information is given to us so we can solve for the bodyโ€™s acceleration when itโ€™s on the plane. Weโ€™ll need to know that to answer this question of how far the body moves up the plane projected at an initial velocity before coming to rest.

To solve then for this bodyโ€™s acceleration as it slides down the plane, itโ€™s important to realize that this acceleration is constant. If we call the angle of inclination of our smooth plane ๐œƒ, then ๐‘Ž, the acceleration, is equal to ๐‘” times the sin of ๐œƒ. The point here is that acceleration is constant because ๐‘” and ๐œƒ are constant. Therefore, we can calculate the acceleration of our body using this given information by referring to the equations of motion. We wonโ€™t list all four of those equations. Here, weโ€™ll just write down one. This equation of motion says that the displacement of a body is equal to its initial velocity times the time elapsed plus one-half its acceleration times that elapsed time squared.

Now, in our case, the distance traveled is ๐‘‘, and the change in time is ฮ”๐‘ก. And because our body is released from rest, that means that ๐‘ฃ sub ๐‘– in this equation, the bodyโ€™s initial velocity, is zero. Therefore, ๐‘‘ equals one-half ๐‘Ž times ฮ”๐‘ก squared. If we multiply both sides of this equation by two and divide both sides by ฮ”๐‘ก squared, we get an equation where ๐‘Ž is the subject. We wonโ€™t yet solve for ๐‘Ž, but we will come back to this equation in a moment.

For now, letโ€™s consider this second stage of motion for our body. Weโ€™re to imagine itโ€™s projected up the plane at 12.6 meters per second. We know that thanks to the effects of gravity, the body will slow down over time and eventually it will come to a stop just instantaneously.

If we call the distance the body travels as it slows down from 12.6 to zero meters per second capital ๐ท, then we can see that itโ€™s this distance we want to solve for to answer our question. To do this, weโ€™ll once again refer to the equations of motion. To see what that equation is, letโ€™s clear away our problem statement, and weโ€™ll now refer to this equation of motion. It says that the final velocity of an object squared equals its initial velocity squared plus two times its acceleration times its displacement. Written in terms of our variables, we could say that ๐‘ฃ๐‘“ squared equals ๐‘ฃ๐‘– squared plus two times ๐‘Ž times capital ๐ท.

And notice this: because our object ends up at rest, that means that ๐‘ฃ sub ๐‘“ is equal to zero. So we can write that zero equals ๐‘ฃ sub ๐‘– squared plus two ๐‘Ž๐ท. Since itโ€™s ๐ท we want to solve for, letโ€™s rearrange this expression. We get that ๐ท equals negative ๐‘ฃ sub ๐‘– squared over two ๐‘Ž. And we can now substitute in the expression we have for the acceleration ๐‘Ž. Now, regarding this minus sign, if we consider ๐‘ฃ sub ๐‘–, the initial velocity of our body, to be positive, then that means weโ€™re saying that motion up the incline is in the positive direction. Therefore, motion the opposite way is in the negative direction. And thatโ€™s the way our acceleration ๐‘Ž will point.

By this sign convention, ๐‘Ž is a negative quantity, and therefore the negative sign in numerator and denominator will cancel out. Weโ€™re finally ready to plug in for ๐‘ฃ sub ๐‘–, ๐ท, and ฮ”๐‘ก. Leaving out units, we use a value of 12.6 for ๐‘ฃ sub ๐‘–, 25.2 for ๐ท, and four for ฮ”๐‘ก. When we compute this fraction, we actually get a result of exactly 25.2. This distance is in meters. So we can say that if our body is projected up the incline at 12.6 meters per second, the distance that will move before it comes to rest is 25.2 meters.

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