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1- Two forces_ F1 and F2 act On the 7.00 kg block shown in the drawing: The magnitudes of the forces are F1=59.0 N and F2=33.0 N and the coefficient of kinetic fric...

Question

1- Two forces_ F1 and F2 act On the 7.00 kg block shown in the drawing: The magnitudes of the forces are F1=59.0 N and F2=33.0 N and the coefficient of kinetic friction between the block and the ground is 0.05. What is the horizontal acceleration (magnitude and direction) of the block? (15 poiuts)'70.027.00 kg F}3- The astronaut in the drawing is standing OnL scale inside satellite. which rotates in circular orbit (radius to earth center= 6400.0 ki) ATOUd the earth Astronaut $ mass is 80.0

1- Two forces_ F1 and F2 act On the 7.00 kg block shown in the drawing: The magnitudes of the forces are F1=59.0 N and F2=33.0 N and the coefficient of kinetic friction between the block and the ground is 0.05. What is the horizontal acceleration (magnitude and direction) of the block? (15 poiuts) '70.02 7.00 kg F} 3- The astronaut in the drawing is standing OnL scale inside satellite. which rotates in circular orbit (radius to earth center= 6400.0 ki) ATOUd the earth Astronaut $ mass is 80.0 kg and the satellite is moving with constant speed of 450 m/s around the earth on the orbit_ What wonld be the apparent weight of the astronauut shown by the scale? Mass of the earth is ~6x10?! kg and G= 6.67x10 Nm? /kg?_ (20 points) Orbit 3- The drawing shows two boxes (m=3.0 kg and mz=2.0 kg) OH frictionless plane inelined at augle 0 =30" ad 02=60" Tle boxes AFO conected Via cord of negligible IHALSS and the pulley is also [rictionless Ac uassless_ What is the tension in the eord? (15 points)



Answers

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?

In this problem, we're asked to draw a free body diagram and calculate the acceleration for two masses on an incline plane. So we're gonna start with our diagram and these messes air connected over a frictionless pulley. And this has a massive three M, and this has a mess of em. And so for part A, we just want to draw the free body diagram. So we have normal force for particular tour surface mg straight down the ramp, straight down, we have a tension going up the ramp and a friction down the room before mess em. And then for mastery M, we have the attention going up and if three mg going down. And so for part B, when we're asked us offering its acceleration, this is just gonna be in terms of variables, and we're actually gonna treat this is the system. So we're gonna take all the all of the forces that would affect its motion. So if you look at M G, it's gonna have two components. It's gonna have a component that's adjacent and a component that is opposite. And these components MG signed data friction. These tensions and three mg are all going to affect the acceleration. So we're gonna take this direction down for three. We're gonna take that direction is positive. Sobering to say three mg minus T plus T minus F minus mg signed data. And this is Newton's second law. The sum of the forces? No, we're just treating it as a system. So I'm gonna write some of the forces on the system equals mass times, exhilaration, and our mass. Here is our total mass. So that's three em plus m. And then the acceleration is going to be for the system. And here you can notice those those tensions cancel out, and friction can be replaced with the coefficient of friction times my normal force. Now, in this case, my normal force is going to be equal to m g co sign. So friction can be replaced with the coefficient of kinetic friction times, mg coast and data. So replacing all of my variables. With that, we have three m g minus mu m geico sank data minus mg sign fatum all divided by three m plus m, which is four em equals a And my accelerations here can actually counsel. I'm sorry, my masses here can actually counsel. And I'm love with three G minus mu. Geico signed Tatum minus She signed data all divided by four. And then part see asks Well, if we have a coefficient of kinetic friction given as 0.3 and an angle of 45 degrees, how far are we going to travel in three seconds? So for that we can assault for exoneration, given our values. So using 10 for gravity, I end up with three times 10 minus point three times 10 times. He co signed a 45 minus 10 times. He signed a 45 divided by four gives me an acceleration of 5.56 meters per second squared and then, using a cinematic equation, I can assume that the initial velocity was zero since it started from rest and then our distance traveled will be 5.56 times when half times our time squared of about 25 meters

Moment and we have a table if detainer that we measure the mess one mess to at the acceleration and we want to get their mind. I quantity that when plotted horizontal axis of graphs should result in a straight line and make the blood. So with remind the the straight line equation and then we just catch what the but should be. So in this case, we can apply the neutrals like waiting for the system and get M one g manifestation is if also m one a empty. Is it going to end to a We use this t here and to solve for the acceleration and this gives us a is you could took em one over m one plus m two g. So if we plot A versus M one over M one plus m two, this should be a straight line. Where this look is the gravity and we can just catch this plant. We have the straight line here and the values Are there a point to you for maximum? Right here is one. And then or the oh, they're access. We have 0.2 and here we have 0.1 So this is more or less what this blood to look like. And the dots will be something like bear around around it. So we have you have to do a linear fit toe, find a plant or just right and all the straight line where a matches most off the points inside. And if you try to copulate, where is this? The slope were found the Slope IHS 9.9 meet us for a second scrap, which is 1.3% higher than the the graphic that these should be the theoretical value for the slope. But this error is inside. The experimental wears off the experiment.

So here we have the system the free body diagram of of the system we can apply Newton's second law in the X. Direction. For the first mass. This would be equaling two m. sub one times the acceleration in the X direction we have then the tension minus M. Someone G. Sine of theta minus the kinetic frictional force. Equaling M. Sub one times the acceleration in the X. Direction. And so the tension force T. Would be equaling two M. Sub one A. Plus M. Someone G. Sine of theta plus the coefficient of kinetic friction multiplied by the normal force for the Y direction. Some of forces in the Y direction for the first mass here we have then N minus M. Someone G. Co sign of theta. And the acceleration of the wind direction is zero. And so This is equaling zero. This will be equal to zero. And so the normal force and equaling eps of one G. Co sign of data. We can substitute this in. So plug this into here and we have that detention equals M. Sub one A. Plus M. Someone G sign of theater plus the coefficient of kinetic friction. I'm someone G co Cynthia For the second mass. We can apply Newton's second law in the Y direction. This would be equal to M. Sub two times the acceleration in the Y direction. We have M. Sub two G minus detention. Equalling two times the acceleration in the Y. Direction. Or simply a. Because this would be denoted as the acceleration of the system. And so the tension T. It would be equaling two. M. Sub two multiplied by then G minus A. And so from this we can find the acceleration of the blocks. And so we have essentially M. Sub two multiplied by G minus M. Two multiplied by a equaling M. Sub one A plus M. Sub one G. Sine of theta plus the coefficient of kinetic friction. M. Sub one G co sign of data and solving for the acceleration A. This would be equal to the N M. Sub two G minus mm. Someone G multiplied by sine of theta plus the coefficient of kinetic friction multiplied by co sign of data. And this all would be divided by the sum of the masses And so solving the acceleration a after plugging in all of your variables is going to be equal to 2.61 meters per second squared. We can use schematics to find the displacement of block to after 1.51 seconds. So you can say S the displacement. S equaling the initial velocity, multiplied by t plus one half a t squared the initial velocity is zero. So this would simply be equal to one half, multiplied by 2.61 m per second squared, multiplied by then one .51 seconds quantity squared S is going to be equal to then. And this would be well, it depends on your Point of reference. S would be equal to 2.97 m. But this is downwards because again we are taking a look at the second block. So downwards for we can say block two in For Block one, it will be two point it will be 2.97 m up the incline. So if this was ry ac accept rather to access an X axis here, Y axis here for the first block and then for the second block it's simply normal. We can say that then in one direction for why for m sub two would be the X direction for someone. And so This would be 72.97 m downwards. four block to 2.97 m is simply the magnitude of the displacement. That is the end of the solution. Thank you for watching.


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