Video: Speed Increase of a Downward Moving Object

Video Transcript

A wagon sits at the top of a hill. The wagon is given a push that increases its speed negligibly but is just sufficient to set the wagon in motion down a straight slope. The wagon rolls 53.9 meters down the slope, which is inclined 16.5 degrees below the horizontal, and reaches the bottom of the hill. If friction is negligible, what speed is the wagon moving when it reaches the bottom of the hill?

We can call the speed of the wagon when it reaches the bottom of the hill 𝑣 sub 𝑓. And we’ll start on our solution by drawing a diagram of the situation. A wagon sits perched at the top of a long flat hill whose length we’ve called 𝐿. It’s 53.9 meters. The hill is inclined at an angle below the horizontal — we’ve called 𝜃 — given as 16.5 degrees. We’re told the wagon is given a very slight push to set it in motion and then rolls down the hill. And when it reaches the bottom of the hill, we want to solve for its speed, 𝑣 sub 𝑓.

In this scenario, our system consists of the wagon and the hill. And since energy is neither added to nor taken away from this system, we can say that it’s conserved. Applying the general statement of energy conservation to this scenario, we’ll use our freedom to decide when the initial moment and when the final moments are. Choosing the initial moment to be just when the cart is pushed at the top of the hill, and the final moment when it just reaches the bottom of the hill.

We can expand this conservation expression to say that the kinetic energy plus potential energy initially is equal to the kinetic energy plus potential energy finally. At the outset, the wagon is not in motion. So its initial kinetic energy is zero. And if we choose the height at the bottom of the hill to be our reference level of zero, we can say that the final potential energy of the cart is also zero, since it’s at zero height. So the wagon’s initial potential energy is equal to its final kinetic energy.

The wagon’s potential energy is entirely gravitational. And we recall that gravitational potential energy is equal to 𝑚 times 𝑔 times ℎ. Applying that relationship, we write 𝑚 times 𝑔 times the initial height of the wagon equals KE sub 𝑓. We also recall that an object’s kinetic energy equals half its mass times its speed squared. We can write this in for our KE sub 𝑓 expression.

Looking at this equality, we see that the mass of the wagon appears on both sides. So it cancels out. We want to solve for the final speed of the wagon. So we rearrange this equation and see the wagon’s final speed is equal to the square root of two times 𝑔 times ℎ sub 𝑖. The acceleration due to gravity, 𝑔, we take to be exactly 9.8 meters per second squared.

Looking at our diagram, we see that the height of the hill, ℎ sub 𝑖, relates to the hill length, 𝐿, and the angle of inclination, 𝜃. In particular, we can write that ℎ sub 𝑖 is equal to 𝐿 times the sin of 𝜃. So we substitute 𝐿 sin 𝜃 in for ℎ sub 𝑖 in our expression for 𝑣 sub 𝑓.

Looking at this expression, we’re given the length 𝐿, the angle 𝜃, and 𝑔 is a known constant. So we’re ready to plug in and solve for 𝑣 sub 𝑓. Entering these values on our calculator, we find that, to three significant figures, 𝑣 sub 𝑓 is 17.3 meters per second. That’s the speed of the wagon when it reaches the bottom of the hill.