CP A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. (a) What is the tension in each part of the cord?

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Accepted Answer

Welcome back everybody. We have a little bit of a complex system here, but we're just going to draw it out and take it one step at a time. We have that a chocolate box is sitting on a frictionless surface. Now it is attached to a string that passes over some pulley here. Now on the other side of this pulley is, you know, as the string runs over it, we have that an ice bucket is sitting on the bar of the string here. I'm just going to label our chocolate box as C In our ice bucket as B. We are told that the mass of the box of chocolate is one kg and we are told that the mass of the bucket is three kg. We're told also that the radius of the pulley is 20 centimeters or 200.2 m. And initially this ice bucket is all the way up here, right? But it's released, starting at rest, meaning its initial velocity is zero and it reaches a speed of two m per second after one second. Now, we also are told a little bit about the naming conventions here. We're told that this part, this top part of this rope is T. One, and we are told that the bottom part of this rope is T. Two and we are tasked with finding what the tension is in either parts of this rope at this time. Okay, so, before tackling any formulas here, I want to point out some other forces that are potentially at play and that might come into effect here. Now, first off, it's important to highlight that the entire system is moving in this direction, right? So its acceleration is acting downward in this case, look at the chocolate box here, we of course are going to have the force due to gravity which is just the mass of the chocolate box times the acceleration due to gravity and upwards, we're going to have the normal force for this ice bucket outside of the tension. We're just going to have the force due to gravity, which is gonna be the mass of the bucket times the acceleration due to gravity here. Now we are tasked with finding, like I said what each of these tensions are. So T one and T two. Let's go ahead and use Newton's second formula to drive equations for each of these terms. Let's see here, let's look at T one and I'm just going to look in the X direction. Um T one is the only force acting in the X direction. So this is just going to be equal to the mass of our chocolate box times acceleration. Right, Using Newton's force in the Y direction for the ice bucket, we are able to drive that. The tension here is equal to the mass of the bucket times gravity minus acceleration. Now we have it looks like we have all the variables except for this acceleration term. So let's go ahead and calculate that. Using some cinematic formulas here, we know that the final velocity is equal to the initial velocity plus acceleration times time. We were told that it starts out at rest, so this term just disappears. And I'm going to divide by time on both sides. This allows us to isolate the term of acceleration and we get that that is equal to our final velocity divided by our time. We have these values. So let's go ahead and find that that acceleration is equal to two divided by one second, which just gives us an acceleration of two m per second squared. With that we are now able to find T one and T two using our formula. So let's go ahead and do that. That T one is equal to the mass of the chocolate box, which is one times acceleration which is too giving us that T one is equal to two newtons for our attention. Right here we have that, it is equal to the mass of the ice bucket Times are gravitational acceleration minus our acceleration value, which gives us 23.4 Newtons. So now we have found tension one and we have found tension to which corresponds to our answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one

Verified Solution

Key Concepts

Newton's Second Law of Motion

Free Body Diagram

Kinematics