### Video Transcript

In a factory, boxes are transferred between two floors via a smooth inclined plane of length 15 meters and height 12 meters. The boxes are released from rest at the top of the slope and left to slide down freely. Taking π to be equal to 9.8 meters per square second, determine the velocity of a box when it reached the bottom of the plane.

Letβs begin by sketching a diagram of the scenario out. We have our inclined plane shown. It has a length of 15 meters and a height of 12 meters. We might denote the included angle at this end of the triangle to be equal to π. And since weβre working with a right-angled triangle, we can use the Pythagorean theorem to find the length of the missing side in this triangle.

This says that the sum of the squares of the two shorter sides is equal to the square of the longest side. If we denote the missing side to be π meters, our equation is 12 squared plus π squared equals 15 squared. If we work out 12 squared and 15 squared, we get 144 and 225, respectively. And to solve for π, weβre going to subtract 144 from both sides and then take the square root of 81. And we get π is equal to nine or nine meters.

We now imagine the box starts from rest at the top of the slope. We know that the weight of this box is mass times gravity. Thatβs the downward force that the box exerts on the slope. We donβt know the mass of the box. So weβll just call that π.

Weβre now going to list what we know. We know that the initial speed or the initial velocity of each box is zero. We want to find the velocity π£ of the box when it reaches the bottom of the plane. And so weβre going to need to be able to calculate at some point the acceleration. We do know that it travels a distance of 15 meters, so the displacement π is 15. So how do we calculate the acceleration π?

Well, weβre going to use the formula πΉ equals ππ. We know the acceleration is going to act in the direction parallel to the plane as shown. And so before we can use the formula πΉ equals ππ, we need to work out the component of the weight β thatβs ππ β that acts parallel to the plane. And so we can add a right-angled triangle here. The included angle is π.

Weβre looking to work out the component of this force thatβs parallel to the plane. Thatβs the opposite side in the right-angled triangle. And Iβve labeled that π₯ or π₯ newtons. We saw that weβre looking to find the opposite side, and we know the hypotenuse. Or at least we have an expression for the hypotenuse. And so we can use the sine ratio that links opposite and hypotenuse such that sin π is π₯ over ππ.

Multiplying through by ππ, and we get π₯ equals ππ sin π. But in fact, we defined π also to be equal to the included angle in our large triangle. The opposite side in this triangle is 12, and the hypotenuse is 15. So we can say that sin π must be 12 over 15 or four-fifths. And therefore, π₯ is equal to four-fifths ππ. This is the only force we have that acts parallel to the plane.

And so going back to our equation πΉ equals ππ, we replace the force with four-fifths ππ. And we see that four-fifths ππ is equal to ππ. And since the mass of the boxes cannot be zero, weβre able to divide through by π. And we find π, acceleration, is four-fifths π.

We now have everything we need to be able to calculate the final velocity of the box. Weβre going to quote one of our equations of constant acceleration. The equation that links π’, π£, π, and π is π£ squared equals π’ squared plus two ππ . We substitute everything we know, and we get π£ squared equals zero squared plus two times four-fifths π times 15. But remember, we were told to take π to be equal to 9.8. And so evaluating the right-hand side of our calculation, and we get 235.2.

Our last step will be to square root both sides of this equation. The square root of 235.2 is 15.33 and so on. Correct to two decimal places, and we see that the velocity of a box when it reaches the bottom of the plane is 15.34 meters per second.

Whatβs really interesting about this is that this velocity is completely independent of the mass of the box. Any box transferred between two floors will have this final velocity.