Question Video: Finding the Minimum Speed Needed for a Body to Reach the Top of a Rough Inclined Plane Mathematics

A body was being projected up a rough inclined plane of length 300 cm whose highest point is 280 cm from the ground. If the coefficient of friction between the body and the plane was 0.41, find, to the nearest two decimal places, the minimum speed at which the body must be projected to reach the top. Give your answer in centimeters per second. Take the acceleration due to gravity to be ๐‘” = 9.8 m/sยฒ.


Video Transcript

A body was being projected up a rough inclined plane of length 300 centimeters whose highest point is 280 centimeters from the ground. If the coefficient of friction between the body and the plane was 0.41, find, to the nearest two decimal places, the minimum speed at which the body must be projected to reach the top. Give your answer in centimeters per second. Take the acceleration due to gravity to be ๐‘” equals 9.8 meters per second squared.

Okay, so in this scenario, we have a rough inclined plane where the maximum height of the plane is 280 centimeters, while the length along the plane itself is 300 centimeters. We imagine putting a body at the base of this incline and then projecting it upwards in the direction of the incline so that itโ€™s just able to reach the very top of the slope. Knowing that the coefficient of friction, weโ€™ll call it ๐œ‡, between the body and the incline is 0.41 and that the acceleration due to gravity is 9.8 meters per second squared, we want to solve for the minimum speed necessary for this body to be projected up the incline and reach the top. And we notice weโ€™re told to give our answer in units of centimeters per second.

To start working toward our answer, letโ€™s clear some space. And letโ€™s say that the minimum speed, the initial speed of the body weโ€™re trying to solve for, is called ๐‘ฃ sub i. Say that we make another sketch of our body on the incline. This time as itโ€™s in the process of moving up that slope at this instant in time, then the velocity of our body would be in this direction. If we consider the forces acting on the body, we know that thereโ€™s a weight force ๐‘š times ๐‘” acting straight downward. Thereโ€™s a reaction force ๐‘… pointed perpendicularly away from the incline. And lastly, because this is a rough incline, thereโ€™s also a frictional force. Because at this instant our body is moving up the incline and friction always resist the direction of motion, we know that that frictional force โ€” weโ€™ll call it ๐น โ€” points down the incline.

With these forces drawn in, we can now set up a coordinate frame where we say the positive ๐‘ฅ-direction is up the incline and the positive ๐‘ฆ-direction points perpendicularly away from it. We can now use this framework as well as the three forces acting on our body to solve for its acceleration along the incline. Since that will be in the ๐‘ฅ-direction, weโ€™ll consider first the forces along this dimension. We can see that there are two such forces. First, thereโ€™s the frictional force ๐น entirely in the negative ๐‘ฅ-direction. And then thereโ€™s also a component of the weight force. That component looks like this, acting directly down the slope. That component we can see is part of a right triangle of forces. We would like to express this side length then in terms of the length we can call it of the hypotenuse ๐‘š times ๐‘”.

Studying this right triangle, it turns out that this interior angle here is the same as this interior angle on our incline. Those angle measures are always the same. Letโ€™s give this angle a name; say we call it ๐œƒ. In terms of ๐œƒ, this opposite side length of our triangle will be equal to ๐‘š times ๐‘” times the sin of ๐œƒ. So thatโ€™s how we can write the second force in the ๐‘ฅ-direction. These two forces are the only ones acting in this dimension. So, by Newtonโ€™s second law of motion, which says that the net force acting on an object equals that objectโ€™s mass times its acceleration, we can say that negative ๐น minus ๐‘š๐‘” sin ๐œƒ equals our bodyโ€™s mass times its acceleration in this direction.

Now, if we can solve for ๐‘Ž sub ๐‘ฅ, then weโ€™ll be able to solve for the initial speed by which our body must be projected to reach the top. We can note at this point that the frictional force ๐น in this equation can be expressed as a coefficient of friction ๐œ‡ multiplied by the reaction force ๐‘… acting on our body. Knowing this, weโ€™d like to know more about this reaction force ๐‘…. To solve for the value of this variable, letโ€™s consider again the forces acting on our body. In the dimension that weโ€™ve called the ๐‘ฆ-direction, there are two such forces. One is the reaction force ๐‘… in the positive ๐‘ฆ-direction, and the other is the ๐‘ฆ-component of the weight force, which is in the negative direction. Now, just as the ๐‘ฅ-component of the weight force was ๐‘š times ๐‘” times the sin of ๐œƒ, the ๐‘ฆ-component is ๐‘š times ๐‘” times the cos of that angle. Our forces in the ๐‘ฆ-direction then are ๐‘… and negative ๐‘š๐‘” times the cos of ๐œƒ.

And here we can note that in this dimension our body is not accelerating. That is, itโ€™s not moving into the incline or away from its surface. In terms of Newtonโ€™s second law then, we can say that ๐‘Ž is zero in this dimension. And so from this equation, we can add ๐‘š times ๐‘” times the cos of ๐œƒ to both sides, and we find that ๐‘… is equal to that term. This means we can replace ๐‘… in our original force equation with ๐‘š๐‘” times the cos of ๐œƒ, and that gives us this result. Notice in this equation that the mass of our body appears in every term. This means if we divide both sides by our bodyโ€™s mass, that value cancels out entirely. So, interesting fact about our result: itโ€™s independent of mass.

That leaves us with this equation for the acceleration of our body along the slope. Note we can factor out a negative ๐‘” from this expression. Looking at this equation, weโ€™re given the value for ๐‘”, and weโ€™re also given the value for the coefficient of friction ๐œ‡. However, we donโ€™t yet know the cos and sin of this angle ๐œƒ. We can figure these quantities out, though, by considering this right triangle. First, letโ€™s look at the sin of ๐œƒ.

In general, given a right triangle where one of the other interior angles is called ๐œƒ, we can label side lengths of the triangle with respect to that angle ๐œƒ. Thereโ€™s the side opposite that angle, the side adjacent to it, and then the hypotenuse of the triangle. The sin of the angle ๐œƒ equals the ratio of the opposite side length to the hypotenuse length. On the triangle in our example, thatโ€™s equal to 280 centimeters divided by 300 centimeters. We can say then that the sin of ๐œƒ equals 280 over 300. When it comes to the cos of the angle ๐œƒ, that, in general, is equal to the ratio of the adjacent side length to the hypotenuse. In our situation then, it would be the ratio of this side length of our right triangle to 300 centimeters.

If we call this side length ๐‘™, we can solve for it using the Pythagorean theorem. Leaving out units, itโ€™s equal to the square root of 300 squared minus 280 squared or, in other words, the square root of 11,600. Knowing this, we can now write our expression for the cos of ๐œƒ as that square root divided by 300. We can now write out an expression for ๐‘Ž sub ๐‘ฅ, with our values for ๐‘”, ๐œ‡, and the cos and sin of ๐œƒ substituted in.

What weโ€™ll do now is store this result off to the side. But letโ€™s also notice that this value for ๐‘Ž, when we calculate it, will be a constant. There are no variables here, so ๐‘Ž will be constant in time. Thatโ€™s important because it means that the motion of our body on this incline can be described by what are called the equations of motion. There are four such equations describing quantities such as velocity, acceleration, and displacement. These equations can be applied to the motion of any object so long as its acceleration ๐‘Ž in these equations is constant. And as weโ€™ve seen, that is the case for our body moving up this incline but accelerating down the incline.

At this point, our task is to choose which of these four equations of motion is most suited to what we want to solve for and what we already know. We want to solve for the initial speed at which our body must be projected. And we know that our body has this acceleration ๐‘Ž sub ๐‘ฅ as it moves up the incline. And we also know that that incline is 300 centimeters long and that our body has just enough initial speed to reach the top. Since we donโ€™t know anything about the time involved in this objectโ€™s motion, we can eliminate the equations of motion involving that variable. The equation weโ€™ll use then says that the final velocity of an object squared equals its initial velocity squared plus two times its acceleration multiplied by its displacement.

Now, remembering that ๐‘ฃ sub i is the minimum speed necessary for our body to reach the top of the incline, that tells us we can say its velocity at this point ๐‘ฃ sub f is zero. Now, after an instant in time, it may start to slide back down the incline. But the point is there is an instant at which it is at rest. Weโ€™ll let that instant in time be our final instant so that ๐‘ฃ sub f does equal zero. If we apply this equation of motion to our scenario then, weโ€™ll have zero being equal to ๐‘ฃ sub i squared plus two times ๐‘Ž sub ๐‘ฅ times ๐‘ . Here, ๐‘  is the distance along the incline the body needs to travel, in other words, 300 centimeters.

Recalling that itโ€™s ๐‘ฃ sub i that we want to solve for, we can subtract two times ๐‘Ž sub ๐‘ฅ times ๐‘  from both sides and then take the square root of both sides of this equation to arrive at this result. Now, it may seem we have a problem with this negative in the square root sign. We can recall, though, that ๐‘Ž sub ๐‘ฅ itself according to our expression is a negative value. When we substitute that into the square root then, those two negatives will cancel out and weโ€™ll have an overall positive. In order to do this, letโ€™s clear some space and then substitute in our known expressions for ๐‘Ž sub ๐‘ฅ and our known value for ๐‘ , 300 centimeters. Leaving out units, we get this. And notice, indeed, there are two minus signs which cancel one another out.

Before we calculate ๐‘ฃ sub i, we need to be very careful about the units involved. Weโ€™re to give our final answer in centimeters per second. And while our value for ๐‘  is given in length units of centimeters, our value for the acceleration due to gravity is not; that has units of meters per second squared. To make all the units agree properly, weโ€™ll want to convert this to centimeters per second squared. To begin doing that, we can recall the fact that 100 centimeters is equal to one meter. Therefore, to convert ๐‘” to units of centimeters per second squared, weโ€™ll move the decimal place two spots to the right. That is, instead of 9.8 meters per second squared, weโ€™ll have 980 centimeters per second squared.

All the units in this expression are now on the same footing, and we can move ahead with calculating ๐‘ฃ sub i. Rounding our answer to two decimal places, we get 797.09. And as we were careful to ensure the units here are centimeters per second, the minimum speed then at which our body must be projected up the incline so that it reaches the top is 797.09 centimeters per second.

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