A spring is hanging from the ceiling. Attaching a 500 g physics book to the spring causes it to stretch 20 cm in order to come to equilibrium. c. What is the book's maximum speed?

Video transcript

Hey, everyone in this problem. During an experiment, a 125 g object is connected to a light and ideal spring suspended vertically from a rigid support at the equilibrium point. The spring is extended by eight centimeters with respect to its natural length. OK. There's more to this problem, but we're gonna stop there for a minute and just draw out what we have so far. We're gonna work through this problem bit by bit. We're gonna draw just a horizontal line that's gonna be that rigid support. We're drawing it like a ceiling and we have our spring that has this 125 g object attached to it. I'm gonna draw a dotted line and this dotted line represents the natural length of that spring. OK. So without the object attached, this is where the spring would fall to. OK. And we're told that in equilibrium, the spring is extended by eight centimeters. So the distance from this dotted line to the bottom of the spring where we have that object is eight centimeters. All right, let's keep reading this problem and figure out what we need to do. So, a student is asked to stretch the spring down seven centimeters from the equilibrium point. And we are asked to calculate the maximum speed the object could reach. We're given four answer choices here A through D and all of them have the unit of meters per second. A is 0.78, B is 0.9 C is 1.2 and D is 11.8. Let's go back to our diagram for a minute. So we have this initial situation where we have this spring in equilibrium which is eight centimeters from its natural length. And now we have a student who is going to pull that spring. OK. And extend it even further. OK. They are going to stretch it by seven centimeters. So the distance from the object at equilibrium to the object in this stretched state is seven centimeters and we are asked to find that maximum speed. Now let's recall that the maximum speed when we're dealing with a spring like this is V max is equal to Omega multiplied by A where Omega is our angular speed and A is the amplitude of the motion. Now, we don't have any information about the angular speed but recall that we can write Omega as the square root of K divided by M. So we can write V max is equal to the square root of K divided by M multiplied by the amplitude A. Now K is the spring constant. M is A mass. So we know the amplitude of this oscillation. OK. This spring is stretched seven centimeters. And so that's gonna be the amplitude. We know the mass. If we can find the spring constant K, then we will be able to solve for V Max. Now, we're told that the spring constant K is related to the spring force. So let's draw a free body diagram, look at the forces and see if we can calculate that spring constant K. So pointing upwards, we have this spring force in our free body diagram and pointing downwards, we have the force of gravity that is acting on that object. OK. This is a free body diagram for that object. So we have the spring force pointing up force of gravity pointing down and let's take up to be a positive direction. No, let's look at the equilibrium case. OK. And we know that at equilibrium, the sum of the forces is going to be equal to zero. OK? Because the object is not accelerating. What forces do we have? Well, in the positive Y direction, we have the spring force F S in the negative Y direction. We have the force due to gravity F G. And so we get F S minus F G is equal to zero. This tells us that the spring force is equal to the force of gravity. Now the spring force recall is equal to K multiplied by X and the force due to gravity is equal to mass multiplied by the gravitational acceleration G. So we have K X is equal to M G. Remember we're trying to solve for K, so we can use that in our V max equation. So we wanna isolate for K, we can divide both sides by X and we get that K is equal to M multiplied by G divided by X. Now we know all of these things, we know the mass, we know the acceleration due to gravity G and this X value is gonna be that stretch of the object. OK? We have to be a little bit careful here because in the problem, we have two kind of stretches. OK? We have the stretch from the natural length and the equilibrium position. And then we have that additional stretch that the student pulls the object. OK? Remember that we're looking at the equilibrium position here, we have the sum of the forces is equal to zero. And so we're, when we're looking for that stretch X, we're looking for that eight centimeter stretch from the natural length. So K is going to be equal to the mass M. We're given 125 g. We wanna convert this into our standard unit of kilograms. So we take our 125 g, we multiply it by one kg divided by 1000 g. OK? We know that there are 1000 g per every kilogram when we do that the unit of gram divides out. And what we're essentially doing is dividing by 1000 to go from grams to kilograms. We multiply that by the acceleration due to gravity 9. m per second squared and we divide it by that stretch of the spring which we're told is eight centimeters. And again, we want that in our standard unit. So we have eight centimeters, we wanna convert this into meters. So we're gonna multiply and we know that for every one m, there are 100 centimeters. So we multiply by one m divided by 100 centimeters, the unit of centimeters divides out. And what we're doing is essentially dividing by 100. And we can simplify all of this to work out. That K is equal to 0.3125 newtons per meter. In the numerator, we have kilogram meters per second squared that is equivalent to a newton. And in the denominator, we have meters. So we have this spring constant K. Remember that's not what we're looking for though. OK. We need to define that. But what we're really looking for in order to answer this question is that maximum speed the object can reach. So let's remind ourselves of that equation. OK. We wrote that the maximum speed V max is equal to the square root of K divided by M multiplied by A. OK. Now we have everything we need. We get that this is equal to the square root of 0.3125 newtons per liter divided by the mass. OK. Now, the mass is 125 g. Again, we wanna convert to kilograms. We've already shown how to do that. In our equation for K, we had 125 g, we divide by 1000 to get to kilograms. OK? And so when we do that, we get 0.125 kg and then we're multiplying by the amplitude A. Now the amplitude a of the motion is going to be that seven centimeters. OK. That's gonna be that maximum displacement when the student stretches the spring and then lets it go. So we have seven centimeters and again, we wanna convert this to meters. It's gonna be the exact same conversion as when we converted our eight centimeters to meters by dividing by 100. So when we do this for our seven centimeters, we divide by 100 we get that this is equal to 0.7 m. OK? And so our V max equation is now equal to the square root a 15.3125 Newton per meter divided by 0.125 kg. And all of that is multiplied by 0.7 m. And this is gonna give us a V max of 0. 476 m per second. When we're looking at the units inside of the square root we have newtons per meter. Ok. Recall that a newton is a kilogram meter per second squared. So we divide that by meter. We're left with kilogram per second squared. Then we're dividing by kilogram. So we're left with the unit of per second squared. We take the square root and we get the unit per second from that square root. And then we're multiplying that by a meter, we get meter per second exactly what we want for speed. And that is what we were looking for. And we found that maximum speed of this object if we compare this to our answer choices and we are gonna round to two decimal places. We see that this corresponds with answer choice. A 0.78 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.