A 1.5-kg block rests on top of a 7.5-kg block (Fig. 4–70). The...

Question

A 1.5-kg block rests on top of a 7.5-kg block (Fig. 4–70). The cord and pulley have negligible mass, and there is no significant friction anywhere.

(b) What is the tension in the connecting cord?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A 2.5 kg calculator is placed atop a 6.5 kg textbook. The rope and pulley system used to balance them has a neg mass and virtually no friction affects the setup, determine the connecting rope tension if the textbook accelerates the calculator at 2.5 m per second squared. So that's our end goal is we're trying to figure out what the rope tension is if the textbook is accelerated or it's accelerating. So the textbook accelerates the calculator. So if the textbook starts moving at 2.5 m per second squared, and that will be our final answer that we're ultimately trying to solve for is we're trying to figure out what this rope tension value is. So we have our diagram here that's provided to us. And as we can see, we have a green calculator on top of a blue and pink textbook. The textbook is 6.5 kg and the calculator is 2.5 kg and going to the left, we have our force and we have like these black lines connected to a purple POY. So essentially, it looks like the force or the, the vector F force is going to be to the left direction. So that's the direction where the textbook will be accelerated is to the left. Awesome. So with that in mind, let's look at our multiple choice answers to see what our final answer might be. And let's also note that all of our multiple choice answers are in units of Newton's A is 6.3 B is 5.7 C is 4.9 and D is 2.0. Awesome. So first off, let us write down all of our known and unknown variables. So we know that the mass of the calculator, which I'm gonna denote as MC. So this is the mass of the calculator on top of the textbook is 2.5 kg. We also know the mass of the textbook, which I'm gonna denote. The mass of the textbook is M TB is equal to 6.5 kg. And this is the mass of the textbook on the bottom. And we're also told that the acceleration A, which I'm gonna do denote as A is 2.5 m per second squared. And we're ultimately trying to solve for the tension force, which I'm gonna denote as F subscript T. And we do not know what the tension force is right at this moment, this is the final answer that we're ultimately trying to solve for. So in to start solving for this problem, we must first calculate the total force that is required to accelerate both the calculator and the textbook at the same time at a acceleration of 2.5 m per second squared, which we can write this equation as F equals MC plus M TB multiplied by A. So now we can plug in all of our known variables to solve for F or the numerical value for F to be exact. So we know that MC is 2.5 kg kilograms plus 6.5 kg multiplied by 2.5 m per second squared. Which when we plug that into our calculator, we should get when we round the one decimal place 22.5 newtons. So now we must solve for the tension in the rope, which is our final answer. Note that the tension in the rope is the force that is acting on the calculator. We can calculate the tension force by using the mass of the calculator and the given acceleration which we can write as FT equals MC multiplied by A. OK, which we can put a in parentheses too if he wanted to. OK. So now at this point, we can plug in all of our known variables and sulfur the numerical value of FT. So we know that MC is 2.5 kg. And we know that A is equal to 2.5 m per second squared. So when we plug that into our calculator and we round to one decimal place, we will get 6.3 newtons. So now we can verify our answer by solving for the tension force, considering the bottom textbook conditions, we can recall and use the following equation that ft the tension force is equal to the force F minus the mass of the textbook multiplied by the acceleration, which is equal to 22.5 newtons. Because that's the value of F the force that we found together minus 6.5 kg multiplied by our acceleration, which is 2.5 m per second squared. Which when we plug that into our calculator, we will find that FT is equal to 6.3 newtons will me round to one decimal place, which means our final answer must be 6.3 Newton. Since it's been verified. Hooray, we did it. So our attention force has to be 6.3 newtons. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter A 6.3 newtons. Thank you so much for watching. Hopefully, that helped and I can't wait to see in the next video. Bye.

Struggling with Physics?

A 1.5-kg block rests on top of a 7.5-kg block (Fig. 4–70). The cord and pulley have negligible mass, and there is no significant friction anywhere. <IMAGE>
(b) What is the tension in the connecting cord?

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Forces in Connected Systems of Objects practice set

Struggling with Physics?

A 1.5-kg block rests on top of a 7.5-kg block (Fig. 4–70). The cord and pulley have negligible mass, and there is no significant friction anywhere. <IMAGE>(b) What is the tension in the connecting cord?

Watch next

Forces in Connected Systems of Objects practice set

A 1.5-kg block rests on top of a 7.5-kg block (Fig. 4–70). The cord and pulley have negligible mass, and there is no significant friction anywhere. <IMAGE>
(b) What is the tension in the connecting cord?