suppose-you-use-videos-to-analyze-the-motion-of-penguins-as-they-slide-from-rest-down-three-differen

Question

Suppose you use videos to analyze the motion of penguins
as they slide from rest down three different frictionless ramps
$(\mathrm{A}, \mathrm{B}, \mathrm{C})$ and into the water. The distances they travel and the
times elapsed are recorded in the following table:
(a) Find the magnitude of the penguin's acceleration for each
ramp. Assume the acceleration is uniform. (b) Determine the
inclination angle $	heta$ above horizontal for each ramp, assuming that
the accelerations for part (a) are in close agreement with the formula, $a=g \sin 	heta .$ (We will derive this formula in Chapter 5.)

Nathan Silvano · Accepted Answer

So in this problem, we have measurements off Penguins is lighting down from treat different rents into the water. So that&#x27;s written here. The distance and the time Off each ramp we have four point 09 through 0.19 for the ramp A. We have one points and 96 and let&#x27;s see, one wants you to wait for rank B and for upsy we have won 20 right and zero point 66 tree. Okay, so first of all, the problem wants us to discover the magnitude of the acceleration in each room. And how are we going to describe this? We know this is emotion with acceleration, so we can use do secretion. Here is the position as a function of time to describe the position of the of the penguins. So the initial position is here. The initial velocity is also zero. So we&#x27;re going to have X equals GT. That&#x27;s called actually a because it&#x27;s not gravity. Acceleration is just on acceleration. A. The square divided by two. Okay, so for their ramp A, we have acceleration off. Let&#x27;s see, 4.9 times two divided by 2.19 you square. This is it close to one point seven meters per second square. This is the acceleration off the ramp. A Let&#x27;s do the same to the others in order Page. Did your room be We&#x27;re going to have, do you? Acceleration. It grows two nights. See? 1.96 times two divided by 1.8 the square. Okay, so that&#x27;s calculate this. This is it crawls through three point 36 meters her second square and the final ramp. We have acceleration off 1.0 rates times two divided by general 0.663 your 0.66 three square and the final acceleration is going to be four point nine one. You just your second square. Okay. The second item of the problem, uh, problems say&#x27;s that our acceleration is described us g sign of Tita. And we need to discover Tita angle for each one of the accelerations we believe it previously discover. So I just need to do the first deceleration is 1.7. So we going to do for Darren? Hey, going to be 1.7 equals nine point 81. Sign off teeter So sign of Tito. It&#x27;s going to be one 0.7, divided by 9.81. This is a close to let&#x27;s see 0.17 17 tree and the value off teeter. Using a calculator, we can discover that he&#x27;s it&#x27;s nine. Going nine eight, the grease during They&#x27;re saying 20 other too accelerations. We&#x27;re going to have three point 36 it close. Nine point 81. Sign of teacher So sign of teacher secret three point 36 Divided by knifepoint 81. This is 3.36 divided by nine point anyone. Zero point tree for two in the value off Teeter. With this second room, it&#x27;s going to be 20 0.0. Tree the grease and the last one for the third rent. The ramp, See is just 4.91 equals nine point 81. Seen sign off Peter. So sign of Tita is four point. Anyone divided by nine points and you&#x27;re one. This is going to be 0.5. So Tito So Tito is just 30 point zero tree Greece. And that&#x27;s the final answer to the question

João Gabriel Alencar Caribé · Answer

In order to solve this question, we have to discover what is the acceleration of the penguin in the second step off his business. In order to do that, we have to apply Newton&#x27;s second law the second situation as my reference frame. For the second situation, I would choose thes one, a vertical axis and a horizontal axis. I would call the vertical axis. Why, under horizontal Axis X, then applying Newton&#x27;s second law situation to on the horizontal access we got. The following the net force on the horizontal direction is because of the mass off the Penguin Times, its acceleration on the result without action. The net force in the horizontal direction is composed in the second situation on Lee by the frictional force, which points to the left. So minus frictional force is because the mass off the Penguin times he&#x27;s acceleration before the acceleration off the penguin is the question minus frictional force divided by his mass. But in the second situation, the penguin is already moving, so the frictional force is given by the kinetic frictional coefficient times the normal force, then his acceleration next direction is a question minus the kinetic frictional coefficient times the normal force divided by his mass. But on the second situation, it&#x27;s very easy to see that the normal force is exactly compensated by the weight force before the normal force is because of the weight force in these second situation, Then using that normal force is it was the weight force, which is because of the mass times acceleration of gravity. We concluded that his acceleration that&#x27;s direction is he goes to minus the kinetic frictional coefficient times his mass times, the acceleration of gravity divided by his mess. Then we can simplify the masses here to get an expression for his acceleration. So his acceleration is the question minus the kinetic frictional coefficient times the acceleration of gravity. In order to use this equation, we have to discover what is the kinetic friction coefficient. And for that, we have to look at Situation number one, where he&#x27;s sliding down a slope in situation number one. I would choose another reference frame on peak these one, and you call these Axis X and these other access. Why, then, applying Newton&#x27;s second law, that&#x27;s question number one to the horizontal axis so that we can get a relation with the frictional force, we get the following the net force. The X direction is equal to the mass off the Penguin times his acceleration in that direction. But his is lighting with constant velocity instead, number one. Therefore his acceleration is close to zero. Then the net force is equal to zero. But the net force in the X direction is given by two forces the frictional force and the X component off the weight force, which can be decomposed as follows. There is a X component w x on a Y component, though you weigh so the net force can be returning us the weight component x minus. The frictional force on this is the course to zero. Then the wait for us component acts is echoes the frictional force again. The frictional force in the situation is the kinetic frictional force. So it&#x27;s given by the kinetic frictional coefficient times their normal force. Therefore, the weight component acts easy. Of course. True, the kinetic frictional coefficient times the normal force. No, we can start. So see from this figure that if we apply Newton&#x27;s second Law on the Y axis, we will end up in the conclusion that the normal force is equal to the weight forced. Why component? Because he isn&#x27;t moving in these axis. So the acceleration is close to zero. And at the same time, the net force is composed by two forces the normal force and the weight component way. So we conclude that the normal force is equal to the weight component. Why the reform? The weight component acts is equals tomb UK times the weight component. Why? Because the normal force is the cause. That&#x27;s no way to conform it. Why? Damn the kinetic frictional coefficient is the coastal w acts divided by the value. Why? And what is this for that we have to take a look at the triangle that is formed by the weight force on its components. So here we have the weight force year. We have the weight force. Why component? And here we have the weight forced X component. Then we have to discover an angle in this triangle. For that, we can extend the weight forced down here so that it forms a 90 degree angle. So we have a triangle like this. This triangle has a 90 degree angle 6.9 degrees angle and another angle in order for the strangle toe have 180 degrees as a result, off the sun off its intern angles. This I know here must be 83 0.1 degrees at the same time. This angle is ouster 90 degrees, then that in turn angle that we have here it must be such that when you add this angle and a 3.1 degrees, you get 90 degrees. Therefore, this angle it&#x27;s also 6.9 degrees. Then the conclusion is that the angle between the Y component on the weight force is 6.9 degrees. So this angle 6.9 degrees, this is the weight force. Then we can calculate to the tangent off 6.19 reasons follows. It is equals to the opposite site w X, divided by the interests inside the were you right? Then we can use these result in his expression to get that the kinetic frictional coefficient is equals to the tangent off 6.9 degrees. Then now that we know, everybody is the kinetic frictional coefficient, we can go back and calculate what is the acceleration off the penguin in the second step off the situation. So he&#x27;s acceleration is it goes to minus the tangent off 6.9 degrees times the acceleration of gravity, which is approximately 9.8 meters per second squared this give us an acceleration off approximately minus 1.186 meters per second squared. This is the acceleration off the dangling. Then we can now calculate How much time does it take for the pain wind come toe help after entering the second phase off the movement In that second phase of the movement, the velocity as a function of time is given by his initial velocity, plus his acceleration times fainter off time. We know that his final velocity will be close to zero. We also know that his initial velocity is 1.4 meters per second. Then the acceleration is minus 1.186 and then we multiply it by the time it takes to come toe health. So we have to solve this equation for tea. We can do this by sending this term to the other side. So 1.1 it six times t is it goes to 1.4 and then sending this term to the other side, so T is equals to 1.4, divided by 1.186 which results in the time off approximately one 0.2 seconds. And this is the time it takes for the Penguin to stop moving after reaching the second things off the movement.

Averell Hause · Answer

So we&#x27;re first going to consider all the penguins as one system so we can apply Newton&#x27;s second law to the system. The attention of all four penguins together would be equal to the mass of the first, plus the mass of the second, plus the massive very plus the mass of the fourth multiplied by the acceleration of the system itself. And so we can say 222 Newtons would be equal to 12 kilograms plus m sub two plus 15 kilograms plus 20 kilograms multiplied by the acceleration. Eh? Now we can consider Onley penguins three and four so we can say the penguins Whether the tensions of four minus attentions up to this should be only equal to massive three plus massive four times acceleration. And we can say that 111 Newtons Ah would be equal to hear 15 kilograms plus 20 kilograms. Time&#x27;s A and so when we solved for the acceleration, the acceleration would then be equal to 3.2 meters per second squared. We&#x27;re gonna plug this into the previous equation that we have derived and then solve for M two so that M too is found to be 23 kilograms. This would be the mass of the second pigment. That is the end of the solution. Thank you. For what?

Averell Hause · Answer

So this would be the free body diagram for the system that we have here and for we&#x27;re gonna apply Newton&#x27;s second law to both masses. So fair, massive A we have The sum, of course, is in the ex direction. This is going to be equal to force tension minus m sub A g sign of fate US of a. Uh this is equaling M sub eight times A for the master B. We have the sum of forces in the X direction equaling AMSA be times g times sign of fate us that be this is gonna be minus force tension. This will equal m sub b times a And now we&#x27;re gonna add these to a crate to equations Together we can say forced tension minus amps of a G sign of data sub A plus. I&#x27;m so be G sign of status of B minus force of tension. This would equal and some eight times a plus. M c b times a force. Tension, of course, cancels out and we find that the acceleration is gonna be equal to M C B time. Sign of status of B minus mass of a time sign of fate us up A This would all be divided by the total mess. So this would be our answer for part A Now, for the, uh except for part B for the acceleration to be at rest acceleration. So at rest this means that the acceleration must equal zero meters per second squared. Ah, therefore the acceleration would equal. I&#x27;m so be again. Sign of fate us A B minus mass of a sign of fate us of a divided by massive A plus massive p and I forgot for this part. Um, there&#x27;s a factor of G. My apologies Just realized so. So this entire term would be times the acceleration due to gravity. Okay, so here and there will be times G. This would equal zero at this point if this equal zero This means that Massa be time sign of status that be I would be equal two massive a time sign of fate s of a therefore massive B would be equal to massive eh Times Sign of fate us of a divided by sign of fate a sub B This is gonna equal 5.0 kilograms. Uh, most applied by sign of 30 two degrees divided by sign of 23 degrees, and this is equaling 6.8 kilograms. So if the massive block A is equal in five kilograms and the massive block B&#x27;s equaling 6.8 kilograms, the system will be at rest and acceleration would be equal to zero meters per second. If we wanted to find attention here, we can say that for the massive, eh? The force tension minus mass of a times G sign of status of A. This is equaling zero. And so the force tension would simply be equal to the massive A 5.0 kilograms multiplied by the acceleration due to gravity 9.80 meters per second squared times sign of 32 degrees and we find that the forest tension is gonna be 26 Newton&#x27;s if the if the system is at rest and four part the sorry poor for for part c uh, just like in part be, the acceleration will be zero for a constant velocity. So we can say for constant velocity, the acceleration is again zero meters per second squared. Ah, therefore the ah, the same exact relationship applies. However, we can say that the massive a divided by the massive be then equals G sign of Fada, so be divided by G sign of fate. US of a, the G&#x27;s cancel out and we have sign of 23 degrees divided by sign of 32 degrees. This is equal in 0.74 So that would be your answer for part C. That is the end of the solution. Thank you for watching.

Averell Hause · Answer

body diagram for the incline. The angle of the incline would be here. We have a mass on the incline, perpendicular to the surface of contact. Always would be the force normal. And then going straight down would be the weight, the weight mg, And so we&#x27;re gonna choose our courteous and axes. This would be why this would be X. And so we can then say that, uh, in the UAE direction, the sum of forces in the wire direction would equal zero. And this would become the normal force and equaling mg co sign of Fada. We can then say that in the ex direction this would also be zero. And this would be n g sign of Fada equaling the mass times acceleration. And so, after canceling out the mass, we find that the acceleration is equaling G sign of Fada. And that is how we prove the acceleration on a frictionless incline. That is the end of the solution. Thank you for watching

Dading Chen · Answer

Okay, so we know to and go, which is Steve? Bank angle can vehicle to, uh, immersed engine Be square org In his questions, you&#x27;re saying that angle theta you see for the inverse tension 0.5. So now we can say that I should go to here, which is we swirl of RG 0.5. So have we square of our equal to zero point Maggie. And with a cheese, the acceleration of gravity which legal 9.8% Swor So have these grove are, which is the centripetal acceleration is he was 4.9 meters percent. It&#x27;s work, so now has taken together choices from question. So the choice with the closest number two My answer here is C. They&#x27;re 40 Craster, for his question is C and this is the answer for this question.

Problem

At the edge of a roof you throw ball 1 upward wit…

Problem 65 Hard Difficulty

Answer

(a)
(i) Ramp $A : a=1.7 \mathrm{m} / \mathrm{s}^{2}$
(ii) Ramp $\mathrm{B} : a=3.36 \mathrm{m} / \mathrm{s}^{2}$
(iii) Ramp $\mathrm{C} : a=4.91 \mathrm{m} / \mathrm{s}^{2}$
(b)
(i) Ramp $A : \theta=9.98^{\circ}$
(ii) Ramp $B : \theta=20.03^{\circ}$
(iii) Ramp $C : \theta=30.03^{\circ}$

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(a)(i) Ramp $A : a=1.7 \mathrm{m} / \mathrm{s}^{2}$(ii) Ramp $\mathrm{B} : a=3.36 \mathrm{m} / \mathrm{s}^{2}$(iii) Ramp $\mathrm{C} : a=4.91 \mathrm{m} / \mathrm{s}^{2}$(b)(i) Ramp $A : \theta=9.98^{\circ}$(ii) Ramp $B : \theta=20.03^{\circ}$(iii) Ramp $C : \theta=30.03^{\circ}$

Physics

Elyse G.

Christina K.

Marshall S.

Zachary M.

Math Review - Intro

Algebra - Example 1

James S. WalkerPhysics

Elyse G.

Christina K.

Marshall S.

Zachary M.

(a)
(i) Ramp $A : a=1.7 \mathrm{m} / \mathrm{s}^{2}$
(ii) Ramp $\mathrm{B} : a=3.36 \mathrm{m} / \mathrm{s}^{2}$
(iii) Ramp $\mathrm{C} : a=4.91 \mathrm{m} / \mathrm{s}^{2}$
(b)
(i) Ramp $A : \theta=9.98^{\circ}$
(ii) Ramp $B : \theta=20.03^{\circ}$
(iii) Ramp $C : \theta=30.03^{\circ}$

James S. Walker

Physics