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In the figure, small block of mass 017 kg can slide along the frictionless oop- the Oop, with loop radius R = 11 cm. The block is released from rest at point P at h...

Question

In the figure, small block of mass 017 kg can slide along the frictionless oop- the Oop, with loop radius R = 11 cm. The block is released from rest at point P at height SR above the bottom of the OOp. What are the magnitudes of (a) the horizontal component and (B) the vertica ponent of the net force acting on the block at point Q? (c) At what height _ should the block be released from rest so that on the verge losing contact with the track at the tOp of the loop? (On the verge of losing contact

In the figure, small block of mass 017 kg can slide along the frictionless oop- the Oop, with loop radius R = 11 cm. The block is released from rest at point P at height SR above the bottom of the OOp. What are the magnitudes of (a) the horizontal component and (B) the vertica ponent of the net force acting on the block at point Q? (c) At what height _ should the block be released from rest so that on the verge losing contact with the track at the tOp of the loop? (On the verge of losing contact means that the norma force on the block Irom the track has Just then become zero).



Answers

A 10.0 -kg block is released from rest at point ?
in Figure $\mathrm{P} 7.71 .$ The track is frictionless except for the portion between points ? and ? { which has a length of } 6.00 \mathrm{m} The block travels down the track, hits a spring of force constant $2250 \mathrm{N} / \mathrm{m},$ and compresses the spring $0.300 \mathrm{m}$ from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points ? and ? .

All right. So the problem of about energy conservation, but also about forces, uh, so for in part air first, we do energy conservation. So the top you have just potential energy u p know Connecticut and you there. And that's equal to thank you. You have. So that is converted thio a potential on Jack you and can take energia que so m g h will be cool too. Em she our cause That's the height advises to plus, um, kinetic energy is 1/2 Mm. The square. So the times cancel out here to give you a V Her v squared off. Uh, g h minus. Uh, she are times two on then h is five hours. So you have five g r minus gr. So that's two times for gr giving you eight times gr uh, for B squared. Therefore, the force in the ex direction your horizontal force will be do too. The centripetal force and the centripetal force will be directed horizontally. And that's magnitude is given by m d. Squared over r, uh, and so plays v squared here. Um, h e r over r, giving us eight mg for the for the magnitude So you replace values here 0.0, uh, 32 kilograms is mass, 9.8 meters per second. Squared is velocity eyes, the gravitation acceleration. Excuse me. Therefore books F sub X is 2.5 Newton's. That's the magnitude and it is directed to the left because that's the direction off acceleration due to the centripetal force. So the ball will be feeling the fourth and that force to the left of 2.5 minutes. Part B is much simpler because the Net vertical force is just gravity. So it's mg 0.32 times 9.8 meters per second squared, giving you 0.31 mutants as the net vertical force. And it will be it will be directed. The net vertical force will be directed downwards, of course. So in part see, we we want to equate the centripetal force to the gravitational force. This means that M v squared revealing the final velocity over our is equal to M g. So these final final velocity will be quality two times are but L C squared. Therefore now we do energy conservation again kinetic energy at people's potential. India P is equal to final kinetic energy plus final potential energy meaning. But MGH plus, uh, plus zero will be equal to 1/2 and the final squared, which is gr plus 1/2 lips plus, um, excuse me plus m g times to our cause. That's the height rises to a P. And so you get Ah, height of 2.5 times are, uh so 2.5 r is 2.5 times 0.0 point 12 meters. That's our So ah, height is 0.3 meters. And finally, to get a graph up a plot like this, uh, you have to, uh you have to find the normal functional form of the normal force a za function off H. And so and so the way to do that is by first looking at, um, the Net force is so, uh, net radio force on radio force at the final squared over are equal two mg plus normal force. So the normal force will be m times the final squared over our minus G by energy conservation. We then have, uh, initially you have m g h. That, uh, will be converted to 1/2 m. V final squared, plus uh, m times two times two are to mgr. Therefore, uh, the final squared will be equal to after the cancellations and the most planes. And what not will be equal to two times g times H minus two are therefore back in, um, the normal force equation. You have efs of Annika ls toe em times to G into age minus two are over our minus g giving you a normal force off to M G H over our minus five times and cheap. And so this is in the form of y equals X plus c. So let me just write that down here. MX plus C why is Epsom n um from the normal Force X is H m is a slope of this line which is two mg over our and sees the intercept, which is negative, five cheap and this is what you get.

According to energy conservation are the initial total Energy Shwe go to really find out of the energy. So therefore, have such energy equation here we just the initial gravitational potential Energy is you go to Lee my Nokia Angie last e r no ah, gravitational potential Energy in the loop. So we know when Ah, a block star rotating in loop the gravity who you go to? The centrifugal force there will have m G's is equal to Embry Square over our And as you can tell, mass m can cancel out And it took about three square is eulogy are so now he's gr subsidy evil we square to have ah, MGH is a good 1/2 mgr uh, plus energy to our as you can tell, uh, mg Jharkand because all and it will give us the high h is you go to won't have our plus two are, which is equally true if I over to our so for part B. So now let's take a look at the situation and a partner of the loop will still apply the sand energy equation. But now the energy question is a study difference since at a bottle over a loop. The highs is as your point. So there's no, uh, gravitational potential energy in a loop right now. Therefore, have the, um, energy equation is I m Jewish. You won't have every square since Now the height is increased due to age. Remember ages ago? Fire which you are. So if we use file with who are substitute palatial have m g two times flowers Who are you with? The Why have every square as you can sell to? Can we cancel out mess semcken Because a lot and this will give us by grs Could do want happy score So therefore we square you see little tens er and the net force off the block in the loop should be go to the normal for us. Ah, minus the gravity off the block. And this is equal centrifugal force because let me draw that We're here. So where blocking is an A button off the loop. Poison was right here The number of forces pointing upward directly to the center of the circle by the gravity He's point down where? Okay, disease, this is grab penises number force And we know centrifugal force is always pulling toward the center. So therefore have m minus MGI is with memories Grow our and we know Um please, we're can be good too. 10 z are therefore have m times tens er over our as you can see how are can because all and that's why we have, um minus m g is the little 10 mg Eventually have the normal force on the block At bottom of the loop is you go to 11 MGI. So for Harbi, I'm sorry, posse should be Parsi year. So now the block is at the top over the loop. So this can be our energy question for this case, which is m g to H is equal to 1/2 and we score plus energy to our things. A nice l travel loop. That means, uh is at a maximum height in a loop, which is two times radius. So remember the highs issue is you go to ah, fire were to has are so we will have. Ah, And if i m g r is you don't want happened. We square plus two mdrv substitute violent to our voyage. It was not run from here and eventually you have to be we square is you could do 60 are So now have the net force in a loop is you go to numerous past grab the things So this is the look So now the block is at the top of the look. As you can tell in the movie yours is pointing downward and a gravity is always point Our, therefore haven't class M g is equal duties and tropical force which is MB square of our and when you re squares grew 60 are so if we used six years have stable, we square eventually we'll have I'm past mg Zulu six energy. Therefore the normal force should be able to buy energy so for party. So now the block thief the loop enters the class action. So when the block enters the class action you know more force on the block Should we go to the gravity of the block? And these are the answers for this question

Initial energy stored in this plane is half X squared equals the final kinetic energy half M V square. So solving this we find X It was v Times Square. Rudolf M over K equals putting the number seen X camped out to the zero for Ford.

Course number 55 a point. A. The initial potential energy is mg at and when the bluff slides town to die spring then frictional force Just some work on it minus were done by the frictional force, which is mu m g into length off the patch. And this remaining energy should be equal to the finalist spring potential energy. Half a X this way. So sub shooting the values masses 10 into 9.8, Just 3 m minus meal into masked and getting into 9.8 and it is 6 m. This is a great toe. How spring Constant destined to 50 and x 0.3 Solving this, we will get the proficient off planet a correction. You 0.3 doh seven it


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