Motion Along Curved Paths - Online Tutor, Practice Problems & Exam Prep

To determine the minimum speed required for a block to reach the top of a 20-meter hill, energy conservation is essential. The equation used is ${\frac{1/2}{m}}^v=mgy$. Here, the final kinetic energy is zero at the peak, leading to the minimum speed formula: $v=\sqrt{2gy}$. Substituting values gives approximately 19.8 m/s, illustrating the importance of gravitational potential energy in curved path problems.

Curved Paths & Energy Conservation

Hey guys. So for this video, I want to introduce you to a type of problem I like to call the curved path problem. Let's check out the example and we'll come back here and fill out the rest. I've got this block that is traveling with some speed, and I want to figure out what is the minimum speed, what's this minimum v here, so that this block can travel up this curvy path and reach the top of the hill, just barely reach the top. So how do we solve this?

We have the height of the hill. We know this height is equal to 20. So how would I solve this? Do I use forces? Do I use kinematics? Well, there are a couple of reasons why those approaches aren't going to work. For one, we actually don't have the angle of this slope because it's constantly changing all the time. And even if we did actually know the angle, what's going to happen is imagine you had a box that was sort of at this point right here, the mg would point downwards and your component of mg that's pulling you down would point in this direction. But then when you reach the steeper part right here, the component that's pulling you down is actually in a different direction. So along the curvy path, your acceleration is never going to be constant. So you can't use forces or kinematics. What I want to point out here is that you're always going to solve these curvy path problems by using energy conservation. There's no other way that you can solve these kinds of problems. Now the reason I say only is that there's technically, in some very rare cases, you may be able to solve these using calculus. But almost always, you're going to use them using energy conservation. So let's go ahead and get to this.

So we have our diagram here. Basically, we're trying to figure out from point A to point B what is the speed that we need so we can reach the top of this 20-meter hill. So now we're just going to go ahead and write our energy conservation equation. So this is going to be Ki+Ui+workdonebynonconservative=Kf+Uf. So let's go ahead and eliminate and expand all the terms. We do have some kinetic energy initially. That's really just going to be related to this speed here. So do we have any gravitational potential? Well, here what happens is if your height is 20, then you can say that your height's initial is equal to 0 when you're at the ground here, so you don't have any gravitational potential.

So do we have any work done by non-conservative forces? Remember, that is work done by applied forces or work done by you or work done by friction. And you actually have neither one of those because you're basically just, you know, this block is just traveling, and there's going to be no friction because it's on a smooth hill. So there's going to be no work done by non-conservative forces. Now do we have any kinetic energy final? The whole idea is that you might be thinking that there might be some kinetic energy because when it reaches the top just barely, it might be moving with some tiny velocity like this. But the whole idea behind these problems is the final velocity and the final kinetic energy is going to be 0 because it's kind of related to the idea of the minimum speed.

So, for example, imagine we had a velocity here of 20 or a minimum speed of 20, and then by the time we reach the top, the block had some speed v_b and that was equal to 5. Well, if it still has some speed here, then that means that you could have actually had a smaller initial speed. This could have been less than 20 and you would have, you know, and you still would have made it to the top. Now imagine I had this as 18 and then v_b is equal to 2. You still could have gone a little bit slower to reach the top of the hill. Eventually, what happens is you're going to reach some number here and that's going to be the minimum speed. If you go any lower or any slower than that, you're actually not going to reach the top of the hill and then you're going to come crashing right back down again. So that's the whole idea, is at the top of the hill, we're going to have 0 kinetic energy.

Finally, we're also going to have some gravitational potential energy because we're at some height y_b. So really, we're just going to expand these last two terms here. So we have ¼m v_a²=mgy_b. What we can do is we can cancel out these masses here, and we can go ahead and solve for this v_a. So if you go ahead and rearrange for this, you're going to get v_a=2g y_b. This is really just the square root of 2 9.8 × 20. If you go ahead and work this out on your calculator, you're going to get 19.8 meters per second.

So there's no other way you could have solved this instead of using energy conservation. That's it for this one, guys. Let me know if you have any questions.

Here’s what students ask on this topic:

What is the minimum speed required for a block to reach the top of a hill using energy conservation?

To determine the minimum speed required for a block to reach the top of a hill, you use the principle of energy conservation. The key equation is:

$\frac{1}{2}\mathrm{mv}2=\mathrm{mgy}$

Here, m is the mass of the block, v is the initial speed, g is the acceleration due to gravity (9.8 m/s²), and y is the height of the hill. At the top of the hill, the kinetic energy is zero, so the equation simplifies to:

$v=\sqrt{2\mathrm{gy}}$

Substituting the values, v = √(2 * 9.8 * 20) ≈ 19.8 m/s. This is the minimum speed required.

Why can't forces or kinematics be used to solve curved path problems?

For curved path problems, forces and kinematics are not effective because the angle of the slope changes continuously, leading to non-constant acceleration. This variability makes it difficult to apply Newton's laws or kinematic equations directly. Instead, energy conservation is used because it accounts for the total energy changes without needing to consider the varying forces and angles. By focusing on the initial and final energy states, you can solve the problem more straightforwardly.

How do you apply energy conservation to solve curved path problems?

To apply energy conservation to curved path problems, you set up an energy conservation equation:

$K$_initial + U_initial + W_{non-conservative} = K_final + U_final

Here, K is kinetic energy, U is potential energy, and W is work done by non-conservative forces. For a frictionless path, W is zero. You then expand the terms and solve for the unknown variable, typically the initial speed. This method simplifies the problem by focusing on energy changes rather than forces and accelerations.

What is the significance of the final kinetic energy being zero in curved path problems?

The final kinetic energy being zero in curved path problems signifies that the object just barely reaches the top of the hill. This condition is crucial for determining the minimum initial speed required. If the object had any remaining kinetic energy at the top, it means it could have started with a lower initial speed. Therefore, setting the final kinetic energy to zero ensures you find the minimum speed needed to reach the peak without any excess energy.

How do you calculate the minimum speed for a block to reach the top of a 20-meter hill?

To calculate the minimum speed for a block to reach the top of a 20-meter hill, use the energy conservation equation:

$\frac{1}{2}\mathrm{mv}2=\mathrm{mgy}$

Here, m is the mass of the block, v is the initial speed, g is the acceleration due to gravity (9.8 m/s²), and y is the height of the hill (20 meters). Simplifying, you get:

$v=\sqrt{2\mathrm{gy}}$

Substituting the values, v = √(2 * 9.8 * 20) ≈ 19.8 m/s. This is the minimum speed required.