(III) A 4.0-kg block is stacked on top of a 12.0-kg block, whi...

Question

(III) A 4.0-kg block is stacked on top of a 12.0-kg block, which is accelerating along a horizontal table at a = 5.2m/s² (Fig. 5–43). Let μₖ = μₛ = μ .

(d) What is the force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the table is frictionless?

Accepted Answer

Welcome back. Everyone in this problem. A steel crate with a mass of 200 kg is placed on top of a large aluminum platform with a mass of 500 kg. The platform is moving along a frictionless factory floor. The coefficient of kinetic friction and static friction between the steel crate and the aluminum platform is the same and devoted denoted by a mu. An external force is applied to the aluminum platform to move both the platform and create at a constant acceleration of 2.5 m per second squared. First, we want to determine the force that must be applied to the aluminum platform without causing the steel crate to slide on the platform. And second in B if the coefficient of friction between the steel crate and the aluminum platform is reduced to half of its original value, determine the force that must be applied to keep the platform accelerating at 2.5 m per second squared for our answers. A says that the first force is 1800 newtons and the second force is 1600 newtons. B that is 715 100 newtons respectively. C 601,400 newtons respectively, and D 1800 newtons and 1500 newtons respectively. Now here, let's start with our first case, the case in which the crate does not slide. Now, if we're trying to figure out a force that must be applied and we're talking about motion. Let's ask ourselves, what do we know about force and motion? Well, recall, OK, that force is directly proportional to acceleration by Newton's second law which tells us that force equals mass multiplied by acceleration. So in our first problem here, given the situation where we say my apologies, given the situation here that the force must be applied without causing the steel crate to slide on the platform. OK? Then that means that the total force required to accelerate both the crate and the platform together fa is going to be equal to their total mass. That is the mass of the crate plus the mass of the platform all multiplied by the acceleration. Well, we know that the force applied to move both the platform and the rate is supposed to be applied at a constant acceleration of 2.5 m per second squared. So really then that force is going to be equal to 200 kg plus 500 kg. OK? All multiplied by 2.5 m per second squared. Now, when we put this into a calculator, we should get 1750 newtons. And if we write it to two significant figures like the rest of our answers. That means it's going to be approximately 1800 newtons. So our first force is 1800 newtons. Let's move on to our second condition. OK. So let's move on to B now. What do we know here? Well, in this problem, it says the coefficient of friction between the steel crate and the platform is reduced to half of it, half of its original value. OK. And we want to determine the force to keep the platform accelerating at 2.5 m per second squared. So we need to consider the new kinetic friction acting on the crate. Now again, recall recall that friction if in this case, this is a common F here, we call that friction F is equal to the coefficient of friction multiplied by the normal. OK, multiplied by the normal force. That's mu N. Now, if we think about our create and platform under these conditions, if we think about the forces that are acting on the both of them, OK, then we know that this force required is acting in this direction. FB while the frictional force is acting against that force and this is a kinetic friction. So we call that F prime K. OK. And we know that our acceleration here, OK, is still 2.5 m per second squared. OK. Four or 200 kg and 500 kg weight. Now, since we're talking about a co a kinetic friction here. That means that our co or kinetic friction is going to be equal to the coefficient of kinetic friction mu K multiplied by the mass of the cred multiplied by the acceleration due to gravity. But what do we know? We'll recall that we said that it's now reduced to half the original value of the coefficient of friction. So all of this is actually equal to a half of mu multiplied by the mass of the crete multiplied by G. OK. No, to solve for the value of mu, we'll use the available information from our first case because note that initially the net force acting on the crate is just the static friction. So what that tells us then is that from case A, I'll keep it in red here from case A, then mu multiplied by the mass of the crete multiplied by G is equal to the mass of the crate multiplied by A. It's, it's, it's, it's our static friction here. Now, if we get rid of the mass of recruit, then that tells us that Muji Mui equals A. And thus, Mu let me go to the next line here. This means that Mu is going to be equal to a divided by G which is 2.5 m per second squared, divided by 9.8 m per second squared. And when we put that into our calculator, we get mu or a coefficient of friction to be 0.255 know that we have the value of mu. OK, then we can put it into our formula for our coefficient of kinetic friction to figure out what that force is. And thus figure out the force FB that we need to, to, to, to keep our constant acceleration of 2.5 m per second squared. So with that idea, then OK, this means that our frictional force is going to be equal to a half a half of mu 0.255 or kinetic friction is a half of was reduced to half the original value multiplied by the mass of the crate 200 kg multiplied by the acceleration due to gravity which we took as 9.8 m per second squared. And when we substitute this into our formula, we get our kinetic friction to be equal to 249.9 new types. Now this is pretty good because if we apply this to our diagram, OK? To figure out FB, OK. Then again by Newton's second law, eh buy Newton's second law, then the sum of the forces OK? Which is FB minus FK is going to be equal to the mass of the platform MP multiplied by the acceleration. OK. Now, if that's the case, that means FB, the force needed in case B is going to be equal to FK plus MP. And since we know all of those values, that means FB is going to be equal to FK 249.9 newtons plus the mass of the platform 500 kg multiplied by the acceleration. We want to keep constant 2.5 m per second squared. Now, if we go ahead and put this into our calculator, we should find that we get that force to be equal to 1499.9 newtons, which again written to two significant figures is really 1500 newtons. Therefore, the force we need in case A was 1800 newtons and in case B would have been 1500 newtons. If we look back at our answer choices, that must mean D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Struggling with Physics?

(III) A 4.0-kg block is stacked on top of a 12.0-kg block, which is accelerating along a horizontal table at a = 5.2m/s² (Fig. 5–43). Let μₖ = μₛ = μ . <IMAGE>
(d) What is the force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the table is frictionless?

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Struggling with Physics?

(III) A 4.0-kg block is stacked on top of a 12.0-kg block, which is accelerating along a horizontal table at a = 5.2m/s² (Fig. 5–43). Let μₖ = μₛ = μ . <IMAGE>(d) What is the force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the table is frictionless?

Watch next

Stacked Blocks practice set

(III) A 4.0-kg block is stacked on top of a 12.0-kg block, which is accelerating along a horizontal table at a = 5.2m/s² (Fig. 5–43). Let μₖ = μₛ = μ . <IMAGE>
(d) What is the force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the table is frictionless?