https://youtu.be/bDmw-ydAnfM
PHYS 1101: Lecture Fifteen, Part Five
Okay. Let me walk you through using these problem solving steps in the context of a fairly simple example. We have a 5-kilogram box that starts from rest. It’s pulled across a frictionless floor with a rope, at an angle of 30 degrees. The tension in the rope is 50 Newtons. How fast is it going after it’s moved 2 meters?
First, let me highlight, again, the references to speed. Starting from rest, we’re asked just a speed, how fast it’s going, and we’ve got position information. So again, position and speed let it bring to mind the possibility of using energy conservation to solve the problem. That means our basic equation we want to start with is this one, which defines the constraint of energy going into this object, and what that means in terms of the speed change, which is the right side of the equation.
So, here are steps. As always, what object do we need to focus on? That’s the box, in this case. That’s the object that’s undergoing this motion, that’s being pulled around. In order to work the left side, I need to think about all the forces, from start to finish, on this box, and the work that they do. So, I want to have a little sketch of the scene. Here’s my box. Here’s my displacement vector. It’s being pulled to the right for 2 meters. Here are the forces I’ve got sketched, that are on the box. I know I always have weight. Then, I have to go around the box and see what else is in contact.
Here, I’ve got a rope touching the box. A rope pulls. That’s responsible for a tension force. As I continue around the box, when I hit the surface that’s also in contact, I know I have a normal force. The surface pushes up. I’ve drawn my normal force a little smaller because I visually just appreciate that the tension has an upward component, in addition to a rightward component. I have no acceleration up/down. So, the sum of everything up will have to balance the weight that’s down. So, this roughly looks like the right length of forces to ensure that that’s true.
I know that the object’s on a frictionless floor. So, I don’t have to worry about friction along this surface. I don’t have a friction force to add. So, this object is going to accelerate to the right because I have a net force to the right. So, I might sketch in here, just a little a to be sure I have a good picture. This object’s going to speed up. Starts from rest, and it’s going to be going faster by the time it’s moved this 2 meters.
So, to do the work, let’s think about just the left side and what the different contributions are going to be. That net work, I need to consider the work done by every force. I’m going to just explicitly write them all. The work that each of these does is going to be force times distance, but it’s the part of the force in the direction of motion.
As I pointed out above, any force that’s perpendicular to the motion does 0 work. It does not help speed up or slow down the object. I’ve got two forces here that are perpendicular, the normal force and the force due to gravity. Right away, I note that those two contributions are zero. They do no work on the object. So, I only end up with this one contribution. My net work is going to be only the work that gets done by this tension force.
So, I can either now break that down and write out what that would be, or I could do it later. I guess I’ll do a quick sketch of it now. Let me draw. I know my tension is at an angle. To calculate the work, I want to draw that force beside the displacement vector like hands on a clock, so the two tails together. Then, my angle, I’m going to sketch on here, is 30 degrees. The part of the force that does work is this part. That effect is captured by writing the work that the tension does.
What’s my recipe? It’s the magnitude of the force times the cosine of the angle, when you draw them like this, tails together like hands on a clock, times the magnitude of the displacement vector, the distance that it’s pulled. So, this is what I will substitute in for the net work. Ultimately, that’s what is going to get substituted in for my fundamental equation that I’m putting this information into.
Now, let’s think about the right side. The right side of my basic equation is kinetic energy final minus kinetic energy initial. Each of these, in turn, is the product of one-half times the mass times the speed at each of these instances squared. Meaning, the first term there turns into one-half mvf2, and the last term turns into one-half mv02.
So, to plug in these numbers, I would need to know the mass for the object, which I do. It’s a 5-kilogram box. Let me just think about what these variables vf and v0 are. They are the speeds at this initial instant, at the start, and at the finish. At those two instances, what is v0? And what is vf final? Do I have that information or do I know it? Well, I’m told if you scroll back and reread the problem, that the book or the box starts from rest. So, that tells me my initial speed is 0. I want to know how fast it’s going after it’s moved 2 meters, which means I’m trying to solve for this variable. That’s what I want.
So, that’s enough for now, thinking about the right side. Let’s make our clear list of knowns and that variable we want to focus on.
Okay. If you scroll back and read the problem, you see that we are told the mass of the box is 5 kilograms. We know that the tension in the rope, which is the magnitude of the tension force, is 50 Newtons. We know the v0 speed is 0, as I already indicated. The angle information there, I already have drawn on my picture. I’m not going to repeat that here. I also know that the displacement vector has a magnitude of 2 meters. The variable I want is vf. This is the variable that’s going to show up in my fundamental equation that I’m going to do my algebra to focus on and to isolate. So, let’s do that.
We’re now ready to work with our basic equation, this fundamental equation that represents this energy conservation. So, let’s start pulling in the information that I’ve learned needs to go into this equation. Let me put it down here. So, for the left side, I know that for the net work, all I had, all I was left with was the work that the tension force does. I’m going to substitute that in. On the right hand side, let me start substituting in the detailed expression for kinetic energy, the one-half mvf2 minus one-half mv02.
Let’s see. Well, one simplification I can make at this point is noting that my initial speed has a value of 0. Therefore, this whole last term goes to 0. That’s a special simplification for this problem. Let’s see. I can do a little bit more work on the left side here. I worked out above that the work that tension does is T times the cosine of 30 times s. Paste that in. Erase that highlight we don’t really need. So, I’m substituting that in to my main equation, continuing my customization.
Okay. Let’s see where we are now. I have a number for the tension, a number for the distance, a number for the mass. I’ve already taken into account my initial speed. I know that the variable I want is right there, vf, the final speed. So, I’m set. Just have to do my algebra. I’m going to rearrange. Solve for vf. You can check my algebra on your own, but the final speed is going to be equal to 2 times T cosine of 30 degrees times s divided by m, and it’s then the square root of that quantity.
Plug in the numbers for that. I have 2 times 50 Newtons, cosine of 30 degrees times 2 meters. I have to divide that by 5 kilograms. Then, square root it, and when you multiply all that out, you end up with 5.9 meters per second.
Let me, off to the side here, do a quick unit check for you to see if the units are consistent. We do need to end up with meters per second, and what do we have? If I focus just on the units, I’ve got a Newton in the numerator, meter divided by a kilogram. Let me break the Newton up into its parts. A Newton is a kilogram meter per second squared. That’s just for this part. I still have another meter in the numerator, and then a kilogram in the denominator. So, the kilogram simplifies, and I end up with meters squared over seconds squared, which, yes. When I take the square root of that, I end up with just meters per second.
So, that looks good. That’s about 6 meters per second. That’s about 12 miles an hour. That could be pulling with a tension of about, well, pulling with this 50 Newton force. I don’t know. That’s probably reasonable.