A light rope is attached to a block with mass 4.00 kg that rests ...

Question

A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 15.0 N. (d) How does the tension compare to the weight of the hanging block?

Accepted Answer

Welcome back everybody we have sitting on a table, a crate and we are told that this crate pass of five kg. No, this rate has a cable attached to it which then passes over a pulley. It is passing over the fully right here and it is connected to another crate. But we do not know the weight of this crate. So I'm just going to market with a W right now actually, sorry, apologies, technically, the force acting downwards, is it weight due to its mass times gravity, which we don't know right now because we don't know its mass. So writing out a couple other forces here acting on these two crates, we know that there's going to be a tension force. So I'm just gonna write attention and we are also going to have our for student gravity right here and our normal force right here. Now we are tasked with finding attention as a function of weight of this crate or compared to the weight of this crate, which I'm going to put that way. Well, once these crates are released, these crates or this creates gonna move downward and this creates gonna move to the right meaning that we are going to have an acceleration going around this cable just like that, meaning that our acceleration right here is equal to this acceleration right here. Now let's find that acceleration which will allow us to find the mass of this crate which will allow us to find the weight. So then we can compare to the tension because we are told that the tension upon release of these crates is newtons. So let's go and find that acceleration first. Using Newton's second law, we have that the sum of all forces in the direction we're going to look at the X direction this time up top equal to a mass times acceleration. So plugging in some values here we have that attention is equal to five times our acceleration which is a total acceleration, meaning that our acceleration is equal to 20 divided by five. Making our acceleration four m per second squared great. Now that we have that, let's go ahead and apply Newton's second law in the Y direction. Now to find the mass of our crate that we don't know some of all forces in the Y direction is equal to mass times acceleration. This means that um just for convention sake, I'm actually gonna have the downward direction. Br positive, since we are moving downward, we have that mg minus our attention is equal to our mass times acceleration, isolating our mass. Here we get that are masses, Our attention divided by our acceleration due to gravity minus our acceleration. And we know all these values so let's plug them in. We have that are mass is equal to 20 divided by 9.8 -4 which is 3.45. Now that we have that, let's go ahead and find the weight of our crate weight is equal to our mass 3.45 times our acceleration due to gravity of 9.8 giving us a weight of 33.81. So we know that tension is 20 And we know that the tension is or we are comparing the tension to the weight of this thing. Right? So let's plug in some values. We have that 20 is equal to our mass times gravity, which we know is 33.81. And we're really solving for this variable X. What is that proportionality constant dividing both sides by 33.81? We get that our X is equal to 0.59, meaning that our tension relates to our weight As being 0.59 times our weight corresponding to our final answer choice of B. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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