Question Video: Finding the Velocity of a Box Sliding from the Top of a Rough Inclined Plane Mathematics

In a factory, boxes are transferred from one area to another via a rough inclined slope of length 13 m and height 12 m. The boxes are released from rest at the top of the slope and left to slide down freely. Given that the coefficient of friction between the plane and a box is 0.27, find the speed of a box when it reaches the bottom of the slope rounded to two decimal places. Take the acceleration due to gravity 𝑔 = 9.8 m/sΒ².

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Video Transcript

In a factory, boxes are transferred from one area to another via a rough inclined slope of length 13 meters and height 12 meters. The boxes are released from rest at the top of the slope and left to slide down freely. Given that the coefficient of friction between the plane and a box is 0.27, find the speed of a box when it reaches the bottom of the slope rounded to two decimal places. Take the acceleration due to gravity 𝑔 to equal 9.8 meters per second squared.

Alright, so let’s say that this is the slope that the boxes in this factory slide down. Starting at the top then, initially at rest, we’re told that they slide down this slope and during this time experience a frictional force. The slope is 13 meters long, and the boxes drop through a height of 12 meters under the influence of the force of gravity, with a given gravitational acceleration. We want to solve for the speed of a given box when it reaches the bottom of the slope. We can call that speed 𝑣 sub f.

Symbolizing the coefficient of friction using the Greek letter πœ‡, let’s now clear some space and start working on our solution. A key realization for us is that as these boxes slide down the slope, they undergo a constant acceleration. That is, the gravitational force pushing the boxes down the slope and the frictional force resisting this motion are constant as the box moves. Whenever an object experiences constant acceleration, that means a set of equations, called the equations of motion, apply to describe its motion.

In general, there are four such equations, but here we’ll just focus on one of them. This equation tells us that when an object undergoes constant acceleration, its final velocity squared is equal to its initial velocity squared plus two times its acceleration multiplied by its displacement. Applying this relationship to our scenario, we know that it’s 𝑣 sub f we want to solve for. And 𝑣 sub i, the initial velocity of the boxes, is zero. We know this because they start from rest.

If we then take the square root of both sides of the equation, we find that 𝑣 sub f equals the square root of two times π‘Ž times 𝑠. Here, π‘Ž is our box’s constant acceleration and 𝑠 is its displacement as it slides down the slope. We know that displacement. It’s given as 13 meters. But what about the acceleration π‘Ž of our box?

To solve for this quantity, let’s start by considering the forces that act on a box as it slides down the slope. Sketching a zoomed-in view of a box as it descends, we know it experiences a weight force, π‘š times 𝑔; a normal or reaction force, we’ll call it 𝑅; and a friction force, we’ll call 𝐹, that acts in a direction opposite to the box’s motion.

We write all these forces down because, recalling Newton’s second law of motion, which tells us that the net force on an object is equal to its mass times it’s acceleration, we see we can use the forces acting on a box sliding down the slope to solve for its acceleration. To do that, let’s say that the direction down the slope is in the positive π‘₯ and the direction perpendicularly away from the slope is in the positive 𝑦. If we then consider the forces that act in what we’ve called the π‘₯-direction on our descending box, we can see that a component of our weight force, highlighted here, fits this description.

Now, if we go back to our original sketch at the bottom of our screen, we can define an angle in this right triangle. We’ll call it πœƒ. This is equal to the angle between the components of our weight force in this right triangle. We see then that the weight force component in the π‘₯-direction is π‘š times 𝑔 times the sin of πœƒ.

Returning to our original sketch, we see that this is equal to the opposite side of our triangle, 12 meters, divided by its hypotenuse. That is, the sin of this angle πœƒ simplifies to the ratio twelve thirteenths. From this term, we subtract the frictional force 𝐹. And these forces combined this way are equal to our box’s mass times its acceleration in the π‘₯-direction.

We can now recall that, in general, the frictional force 𝐹 acting on an object is equal to the relevant coefficient of friction πœ‡ multiplied by the reaction force acting on the object. We have that coefficient of friction given to us in the problem statement. And looking at our free body diagram, the sketch including all the forces acting on our box, we can see that the magnitude of the reaction force 𝑅 is equal to the component of the weight force acting like this. Mathematically, this is equal to π‘š times 𝑔 times the cos of πœƒ.

And now, looking back once again at our original sketch, we can notice that this right triangle has special side length ratios. With a hypotenuse of 13 and a height of 12, we know that this is a triangle where the other leg has a length of five units, in this case five meters. We point this out because the cos of our angle πœƒ is equal to the adjacent side length divided by the hypotenuse length. That is, the cos of πœƒ equals five thirteenths. So π‘š times 𝑔 times five thirteenths is equal to the magnitude of the reaction force 𝑅. And multiplying this by the coefficient of friction πœ‡ then gives us an expression for the friction force 𝐹.

Our equation for forces in the π‘₯-direction now looks like this. And notice that our box’s mass π‘š is common to every term. Since we’re assuming the box has a nonzero mass, this tells us we can divide both sides of the equation by π‘š and cancel it out. If we then factor the acceleration due to gravity 𝑔 out from both terms on the left-hand side, we find this expression for the acceleration of our descending box π‘Ž.

Knowing this, we can now substitute the left-hand side of our expression in for π‘Ž in our expression for 𝑣 sub f. When we do that and we multiply 13 by two to get 26, then we have this overall expression for the final velocity of our descending box. And now we can substitute in for the acceleration due to gravity 𝑔 and the coefficient of friction πœ‡. When we then enter this expression on our calculator, rounding our answer to two decimal places, we get 14.45. And our answer has units of meters per second. This then is the speed that our boxes attain once they reach the bottom of this incline.

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