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Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.

A small box of mass $m$ is placed on an inclined plane with an angle of incline of $\theta$ . There is a coefficient of kinetic friction $\mu_{\mathrm{k} \text { between the inclined plane and }}$ the small box. The small box is attached to a much heavier box of mass 3$m$ by a pulley system shown below.(a) Draw free body diagrams of both masses, including all of the forces acting on each.(b) Assuming a frictionless, massless pulley, determine the acceleration of the blocks once they are released from rest in terms of $\mu_{k} g,$ and $\theta$(c) If $\mu_{k}=0.3$ and $\theta=45^{\circ},$ what distance is traveled by the blocks 3 $\mathrm{s}$after being released from rest?

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Chapter 13

Practice Test 3

Section 2

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Hi in the given problem theories an inclined plane, The rough inclined plane which is making an angle of tita waved. The horizontal a block of mass M. Has been kept over it which is tied with the help of a thread. The threat is passing over a frictionless. Massless fully and another block of masks. three M. is tied with the help of the threat. Now we have to plot, we have to draw the free body diagrams of both the blocks. So first of all the weight of this block which is kept over the inclined plane will be acting vertically downward like this. You have this is M. G. Its components, there will be two components of this weight if this angle is he, that this angle will also be theater. So the component of the weight perpendicular to the inclined plane will be MG. Cause theta and its component along the inclined plane will be MG sine Tito. Normal normal reaction applied by the inclined plane to this block will be N is equal to MG cost heat up this three and the block of Mastery um will be having a tendency to Slide to move downward under the influence of its weight. three MG. Due to which this block kept on the inclined plane, inclined plane will be having a tendency to move up so our force of friction will be acting on it in a downward direction. There will be a tension in that thread which will remain same in both of its segments, attention teeth. So here these are the free body diagrams of both the bloods like this, which is the answer for the first part of the problem. So free body diagrams off both the blocks are shown in the adjoining figure. And here we will have to show the acceleration also as both the blocks are connected with each other, so the acceleration a will be same in both of these blocks. Now, in the second part of the problem, we have to find that acceleration for which was to fall. We will use free body diagram of the block of mass three M. As this block is under the influence of two forces three M. G and T. And as it is moving down, so three MG should be more. So the net force acting on it will be three and G minus T. And using Newton's second law of motion, this net force should be created by the product of mass with the acceleration. So here it becomes first equation. Now we will use free body diagram of the block kept on the inclined plane which is having a mass of him. And as this block is also under the influence of two forces B upward along the inclined plane and MG. Sine theta And force of friction. So this is under the influence of three forces MG santa and air force of friction downward. A long time climb plane. But as the block is moving up so attention T. Should be more. So the net forces acting on it will be T minus the sum of mg. Scientific to with the force of friction, and that should be equal to M. Into a Again expert Newton's 2nd law of motion. No, as for so friction is given us the coefficient of friction, mu k times the normal reaction and normal reaction here is MG cost theta. So finally this equation comes out to be T minus mg sine tita minus mu UK MG. Because tita is equal to me. That is the question number two. And if we right here, equation one again, that was three M. G minus T is equal to three M. A. And now if we add these two equations, the tension will be cancelled, leaving behind three mg minus mg. Sine tita minus mu k mg because tita is equal to me. So the acceleration here will come out to be canceling this empty from all the terms. The expression for acceleration is 3G -G sign tita. And here, also, as we have added decorations, it will come out to be for me. So this is four M. E. So here this is 3G -G scientist to minus mu K G because tita divided by four. And here it is the answer for the second part of the problem expression for the acceleration of the system. Now, finally, in the third part of the problem, Coefficient of friction has been given a 0.3 Angle Theta has been given us 45° in time. T because 23 seconds we have to find the distance traveled by the object of mastery. M given that it has started from rest. So initially speed, you will be taken as zero so to find it. First of all, you will find acceleration of the system. Using the answer we found in the second part of the problem. So that is three G means G can be taken here as a common out. So this is three minus sine. Theta means signed 45 degree minus mu game, which is 0.3 G. Has been taken as a comin out Cost histamines cause 45°.. Here this is G means 9.8 m per second squared, divided by four. So finally this acceleration comes out to be 5.2 m per second squared. Hence using the second equation of motion as is equal to U. T Plus have 80 square that it stands traveled by. The object will be zero plus half into 5.2 m per 70 square into three square. So finally this distance comes out to be 23.4 Meta, which is the answer for the 3rd part of the problem. Thank you.

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