a-block-of-mass-m-is-at-rest-on-a-frictionless-horizontal-table-placed-in-a-laboratory-on-the-surfac

Question

A block of mass m is at rest on a frictionless, horizontal table placed in a laboratory on the surface of the Earth. An identical block is at rest on a frictionless, horizontal table placed on the surface of the Moon. Let F be the net force necessary to give the Earth-bound block an acceleration of a across the table. Given that g$_{Moon}$ is one-sixth of g$_{Earth}$, the force necessary to give the Moon-bound block the same acceleration a across the table is
(A) F/6
(B) F/3
(C) F
(D) 6F

Bret Rosen · Accepted Answer

in this problem. We have a block on a frictionless horizontal table and we want to know how much force we need to apply on the earth and on the moon to get our block to the same acceleration. So the key comes up here. Is that because there is no friction and it is cheerily horizontal motion. Gravity has no effect. So it will move exactly the same on the Earth and the moon, since friction does involve the force of gravity. But in our our case here, So we have our table with our block of mass and and we&#x27;re applying some force to it. And so it&#x27;s going to be moving, uh, in purely in the horizontal direction with acceleration a So on earth, the emotion is just going to look like yeah, according to F equals m A. And on the moon, we will have the same situation. Uh, people&#x27;s m a. So gravity will not enter into this determination. When we have this motion that&#x27;s purely horizontal. So gravity is acting in a purely vertical direction so orthogonal to this direction. So our force on the earth and the moon is the same. All right, so out of our answer choices, we will choose See f to have the same force to produce the same acceleration on the earth and the movement.

Ze-Han Lee · Answer

So in this problem we have a model like this. We have a cliff, and there is a block over here, and there is a pulley, so it is drugged by another block. All right, so we already know that. Ah Ah, The mess off the small bloke over here is ah, enough to ah make this. Ah, make this a big block move. So Ah, we have four choices and we want to know that. Ah ah, you see, So we already know the acceleration of the system and the message of the blocks. So suppose the mess of the small ones this morning. This is a capital M, and we also looks at a recent right. So for choice, eh? Ah, the net force on each book. This is able to this is able to do that way we can do the calculation, actually, because we already know that separation rain So net force on the net force on each block is, uh ah just equal to so suppose for this one is simply go to m times ain&#x27;t right. And for this one is simply able to end small sometimes say so. This will be the net force on the two blocks. And what choice? Pay attention the stream. So we can also do the calculation. Find it. So suppose we&#x27;re looking at the small small book so we can list equation I&#x27;m g minus tension in this ah industry Iqlim time say right because the for the smaller block Ah, we only have the repetition of force and attention so well is that this equation And we can find that the t equal mg minus I may write and for choice say ah, the coefficient of a kind of friction between the table and the blow and the block on MSN. We can also do the complicated calculation to find Ah, this coefficient Because for the bigger bloke, we have only ah, attention force in the friction. Right, so we can list the question and Sorry, the capital m Time&#x27;s a I&#x27;m sorry. We convinced equation Attention force Minus the friction force, which is mg Sorry the capital M capital M times G times Mu K where? Mu case That kind of friction constant. Okay. Ah, equal capital am time. Say All right. So now that we can find attention force so we can use to the value of distance tension forcing this equation so that we can tend the value of the mu K. All right, and therefore, the last choice. Ah, the speed of the block off mess capital M wide reaches the edge of the table. We&#x27;re not able to find it because we do not know that this is from the ah, from this position to the end of the cliff. All right, Well, only if we know this. Ah, distance. So that so that we can find it a final speed, all right?

Averell Hause · Answer

This would be the quick free by diagrams for the top Lock em So t and the bottom block m sub em Sabi brother were first going to calculate the coefficient of static friction for the surface between the two blocks. We know that the when the force applied is at a maximum. The frictional force between the two blocks must also be at a maximum. So we can say that f sub t 12 Newtons of force has to be applied to the top lock for slipping to take place so we can say f sub t equaling the maximum static frictional force equaling the coefficient of static friction multiplied by the force normal t we can say that. Then the maximum static frictional force is gonna be equaling to the coefficient of static friction multiplied by the mass of block tee times G. We have then that the coefficient of static friction is gonna be equaling. F sub t divided by M sub t g. This is equaling 12 Newtons, divided by 4.0 kilograms, multiplied by 9.8 meters per second squared and this is giving us 0.31 So this would be the coefficient of static friction. Just keep this value in mind, using the same reasoning for the two masses. Together, we can say that then the maximum applied force F would be equaling the coefficient of static friction multiplied by the mass of the top block, plus the mass of the bottom block multiplied by G for part two. A. We can simply solve this. This would be equaling 2.31 multiplied by 4.0 plus 5.0 kilograms multiplied by 9.8 meters per second squared. And this is giving us 27 Newton&#x27;s and then for part B to find the maximum acceleration, weaken, say, maximum acceleration equaling F over the total mass multiple rather equaling the coefficient of static friction multiplied by G. So this is equaling 0.31 multiplied by 9.8 meters per second squared, and we find that the maximum acceleration is equaling 3.0 meters per second squared. This would be our final answer. That is the end of the solution. Thank you for watching

Averell Hause · Answer

So first for party. We have three M three times a certain mass, and on top of this block, we have a block of Mass M. And so we have a force applied and lead going to the right. So drawing the free body diagram for the first block on top, we have the force normal. And then going down with BMG to the right would be the force. And then for these larger mass we have going up would be again forced normal to the left. However, we have force of friction F lower case that further to the right, we have the force applied capital F and then going down to be three mg three times the force of gravity as, ah relative to the smaller mass. So this would be the two free about it. Aaron to free body diagrams for a part. Ay. And we can say for part B, we can say in this case, this system, uh, the we could say the force of friction between mass three em and mm causes allows the block of mass mm to accelerate. And then this would be your answer for piano. For part C, we can finally find the acceleration in terms of the force applied and the mess. So the sum of the forces is going to be equaling, Um plus three m multiplied by a the acceleration and the sum of the forces, actually the only force of the only force f. So in terms of the system, there isn&#x27;t any friction. There&#x27;s on Lee the frictional force. Ah, there&#x27;s no friction between If this is the table, there&#x27;s no friction between the larger mass and the table. There&#x27;s only friction between the larger mass and the smaller mess on top, which is again causing block of a mass M to accelerate. So when we saw for the US acceleration of the system itself, the only forces acting on the entire system is going to be forced is gonna be the force applied. Given that here there isn&#x27;t any friction between the larger mass and the table that it is placed on. So the force applied would be equaling four times a day. And so the acceleration is found to be the force applied, divided by four times the mass. This would be our final answer for part C. That is the end of the solution. Thank you for watching

Averell Hause · Answer

so here for part A. We know that a small segment of the rope has mass and is being pulled down by the gravitational force off the earth. So we can say that rope has mass and therefore has a gravitational force. Given this, we know the equilibrium is reached because neighboring portions of the rope pull sufficiently up on it. So because tension is a force along the rope, we can then say that here at least one of the neighboring portions must slope away from the segment that we&#x27;re considering. So because the tension force is an upwards force or rather hasn&#x27;t upwards component weakened, then say, of course the rope will sag. Four part be weakened, then say that the only force acting with a horizontal component is the applied force. So treating the block and the rope as a single object weaken right. Newton&#x27;s second law and we can say that F is equaling the mass plus, um, times A. So this would be the sum of the mass of the block in the rope times the acceleration. Therefore, the acceleration is simply going to be equal to the X component of the Applied Force. We can say applied X, divided by that mass that some of masses and that would be the formula for the acceleration. Four parts see weakened, then say the force of the rope F R is the only force with a horizontal component acting on the block. So then we can apply Newton&#x27;s second law for on Lee the Block and we can say that the force of the rope with equal the mass of the Loch Times acceleration and this would be equal to the mass of the block times f uh, essentially, to avoid confusion, we can just call this F So the, uh after we calculated in rather that was given in part B and then divided by the total mass of the system, we&#x27;re essentially just plugging in for the acceleration that we have found in Part B. And so we can say that again. The force of the rope would simply be equal to M F over M plus and and finally four part D. We can treat the block and half the rope as a single object. So here we can say m plus 1/2 and would be the total mass where we have a court where the horizontal force on the block and half the rope would be t sub M t&#x27;s that Deb, we would be considered the midpoint the tension at the midpoint of the rope. So we can say that the tension at the midpoint of the rope would be equal to M plus 1/2 mm times acceleration. And then we can plug in, Sir. D would be equal now for part D. T. C M would be equal to m plus plus an ogre, too. Times f divided by the total mass. And we can simplify such. This would be asked times to m plus m divided by two times and plus. And this would be our tension at the midpoint of the rope. That is the end of the solution. Thank you for watching.

Averell Hause · Answer

we have two masses, one with mass m dealing with mass to em. We can apply Newton&#x27;s second law to the system as a whole. So we have the total mass times acceleration. This would give us that the net force where f would be equal to three times the mass, the combined mass times the acceleration or we consult for the acceleration. This would be equal to the force divided by three times the mass three M. Now we know that the force on the second block of Mass M f sub to we could say equals m a. Now we&#x27;re going to for a plug in F. This would be equal to M multiple brother for a plug in F over three m. So it would be f over three m cancelling out the mass M. And therefore we see that f sub to equals f sub three. So the question is asking us what force does the block on the left? What forced us? The left bloc exert on the right block and it will be equal to the force that is being applied to the right block horizontally divided by three. So this would be our final answer. That is the end of the solution. Thank you for watching

Problem

A crate of mass 100 kg is at rest on a horizontal…

Problem 8 Easy Difficulty

Answer

C) F

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