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A spring (k = 800 N/m) is placed in a vertical position with its lower end supported by a horizontal surface (see Figure): The upper end is depressed 20 cm, and a 0...

Question

A spring (k = 800 N/m) is placed in a vertical position with its lower end supported by a horizontal surface (see Figure): The upper end is depressed 20 cm, and a 0.5-kg block is placed on the depressed spring: The system is then released from rest: How far above the point of release will the block rise?Select one:a. 2.4mb. 3.2mc.2.8 md.2.0 m

A spring (k = 800 N/m) is placed in a vertical position with its lower end supported by a horizontal surface (see Figure): The upper end is depressed 20 cm, and a 0.5-kg block is placed on the depressed spring: The system is then released from rest: How far above the point of release will the block rise? Select one: a. 2.4m b. 3.2m c.2.8 m d.2.0 m



Answers

A spring with spring constant $k=620 \mathrm{N} / \mathrm{m}$ is placed in a vertical orientation with its lower end supported by a horizontal surface. The upper end is depressed $25 \mathrm{cm},$ and a block with a weight of 50 $\mathrm{N}$ is placed (unattached) on the depressed spring. The system is then released from rest. Assume that the gravitational potential energy $U_{g}$ of the block is zero at the release point $(y=0)$ and calculate the kinetic energy $K$ of the block for $y$ equal to (a) 0 , (b) $0.050 \mathrm{m},(\mathrm{c}) 0.10 \mathrm{m},(\mathrm{d}) 0.15 \mathrm{m},$ and $(\mathrm{e}) 0.20 \mathrm{m} .$ Also, (f) how far above its point of release does the block rise?

Given Marshall to block two kg block. Start from rest on the frictionless surface, having them initial height. Quite five weekend. Mhm the spring constant of the spring As soon as the figure is 4 15 and no permit. Er we have to fight the compression industry and finding the height of the block. When it go back, find a like then go back airplane Contribution of energy potential energy is called to kinda technology mg By initial it's called a half M B square. So where the city you will get a route to get by Subsidy of the volume G is 9.8 an initial itis 0.5. So it is to be 2.214 m per second. Not applying the contribution of energy. Half MB square is going to have k X squared So compression in the spring you will get I am we square by substitute seven moss after block is given two kg Spain. We have calculated 2.214 and spring constant is 4 50 newtons per meter. So x you will get quite 209 me different? No see part suit Yeah, surface It friction lists so no loss, any kind of technology. Hence final position be required to initial. That is 0.5 m that son. Thanks for watching it.

For this problem on the topic of energy conservation, we're told that a block with mass one kg is compressing a spring, which has a spring constant of 100 newtons per meter, by a distance of 20 centimeters, the spring is released and the block moves across a horizontal frictionless table. It then hits another spring, which has a spring constant of 50 Newton newtons per meter, and compresses that spring. We want to find the total mechanical energy of the system, The speed of the mass while moving between the springs and the maximum compression of the 2nd spring. Now the total mechanical energy can be determined by recalling that. For a conservative system, we have the total mechanical energy equal to the maximum potential energy at some point and the maximum kinetic energy at another point. Now the maximum potential energy can be determined from spring one. So this maximum potential energy is equal to A half times the spring constant of spring, one times its maximum compression X. Max squared. And so the total mechanical energy at the system is equal to a half times The spring constant of 100 newtons per meter for spring one times the maximum compression of the spring, which is 20 times 10 to the minus two m squared, which gives us the maximum what he told you mechanical ingenious system to be to jules. Now, for part B we want to find the maximum speed of the block as it moves between springs. So the maximum kinetic energy K max is equal to the maximum potential energy you max. This implies that M times the maximum speed V max squared Over two is equal to K one times X one maximum squared over two. And so we can rearrange this equation and solve for the maximum speed V. Max. And we get remax to be The spring constant K. one over the mass. M times the Maximum compression for spring one x. max one and that's squared. And so this is equal to The maximum compression of spring one times The Square Root of K. one over M. And so this is 20 Times 10 to the -2 m times the square root Of 100 Newtons. The meta divided by one kg which gives us the speed of the block as it moves between the strings To be two m/s. And lastly for part c. were to find the maximum compression of the 2nd spring. Now we know that uh huh Maximum potential energy on the second spring must equal to the maximum kinetic energy. And so uh huh. K. Two X max two squared where X max two is the maximum compression of the second spring must equal to the maximum kinetic energy. And so rearranging this equation, we get the maximum compression of the second spring X. Max two to be the square root of two times The maximum potential kinetic energy divided by the spring constant K. two. And these values are all known. So this is the square root of two times two jewels, divided by The spring constant of the 2nd spring is 50 newtons per meter. This gives the compression of the second spring to be 2.83 Times 10 to the -1 meters, which is 28.3 centimetres.

So here the given values that were given in this problem and we have to find the velocity of the book at 0.25 meters height were asked to find the distance that spring is compressed when the block runs into it and he has to find a final height that the bloc travels up the ramp. Um, at the end of the problem. So to solve for these speed at 0.25 meters, we're going to use conservation of energy. We know that our initial kinetic and gravitational potential energy is equal to her final gravitational potential and kinetic energies so to solve in Israel MGH and we know the mass of the Bach. It's two kilograms and we know that it is at a height of 0.5 meters. We know that it is at rest at the top of the ramp and on our final position, it is at a height of 0.25 meters and we'll use 1/2 MV squared for kinetic energy. You know, it's a massive two kilograms we're solving for velocity. So when we plug in these numbers, we find that the velocity of the block is 2.21 meters per second. Solve for part B. We're gonna do another conservation of energy equation using this initial gravitational potential. Energy will move that down here zero point five and they're going to set that equal Teoh our potential spring energy which is equal to 1/2 k x squared. And we've been given a K value of 450 Newtons per meter and we're solving for X. So when we plug in these numbers, we find that X delta X The displacement of our spring is equal to 0.209 meters and finally, for part C, we're going to use our potential spring energy to do a final Conservation of energy equations will bring that over here. And we know that Delta X we just solved for is equal to 2.29 meters squared and we're going to set it equal to our final gravitational potential energy of the Bach traveling back up the ramp. And that is going to be two times 9.8 and we're solving for the height. And we know that the point that it stops it has zero velocity, so zero kinetic energy And when we saw this equation for H, we get a height of 0.5 meters, so returns back to its original height, and that makes sense because

High in the given problem. The mass, which is oscillating with the help off the spring on the horizontal platform, is given ence must off the block. MM is equal to 0.400 kg and force constant off. The spring is K is equal to 200 Newtons for meter. No, when the displacement instantaneous displacement off the block from it's mean position is let it be vile, and that is given as 0.1 60 m. Then the speed off the block at that moment is we is equal to 3.0 meter per second. Now, in the first part of the problem, we have to find the maximum speed off the block. Then it will be passing through its mean position. So to find this maximum speed off the block, first of all, using the expression for its instantaneous velocity, which says we is equal toe Amira, the angular, frequency off oscillation square root off a square. Maximum displacement off the block means amplitude off facility emotion minus y squared. And here key is given us am Omega Square. So using that omega squared equals Tokyo. I am, or we can say omega is equal to square root off Gabe, I am so putting this value formula Also here in the expression for instantaneous velocity we get, we is equal to square root off cape I am into square root off a square, minus wise. Where or we can say this is square root off e g Break it a square minus y squared by m No squaring Botha sires we get so we square is equal. Tok is square minus y squared, divided by m muscle the block so plugging in the non values hair for V, the speed off the block that was given us 3 m per second. So that's the square will come out to be nine is equal to four k. This is 200 a means maximum displacement off the block means an amplitude off the oscillator emotion that is not given to us. So it remains as is well minus. Survive means the instantaneous displacement which is 0.16 to the whole square, divided by mass off the block, which is 0.400 kg. So this 200 into a square minus for this 0.1 60 square it comes out to be 0.256 visible to nine in 20.4, which comes out to be 3.6. Or this a square minus 0.256 becomes equal to 3.65 200 which comes out to be 0.18 Hence, a square becomes 010 to 56 plus 0.18 means it comes out to be 0.436 So amplitude off this facility emotion with the square root off 0.0 436 m. Or we can say this is 0.209 m. Hence, the maximum miss speech off the block after its mean position will be given by the product office amplitude with it's angular frequency a Oh, make up what you can say. This is 0.209 for Omega. This is square root off. Gabe. I m k is 200. Bye. Mm. We choose 0.4, so it comes out to be 0.209 Removing this decimal 2000 by four which becomes square root off 500 So finally, this ReMax needs the answer off. First part of the problem comes out to be 4.67 m. But second answer for the first part of the problem, no. In the second part of the problem, we need to find the magazine experience, after which the spring is compressed and that is actually ableto complete your off the social promotion. So maximum dispense. Bye, bitch. The spring is compressed is its amplitude which we have found already and that is 0.209 meter and the third part of the problem actor Maximum compression. We have to find speed off the block and acceleration of the block. So at maximum compression, speed off the block will be zero and its acceleration means capitally will be given by in magnitude. It'll be given by a into Omega Square. So it becomes 0.209 into square off omega, which was square root off 200 by 0.4. So simply it will become hunted by 0.4 square would be removed due to this squaring. So it comes out to be into 500. So this is 0.209 into 500 meter per second square. Or you can say this is 104.5 meter per second squared. Answer for the third part off the problem. Thank you.


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