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A $0.500-$ kg block is released from rest and slides down a frictionless track that begins 2.00 $\mathrm{m}$ above the horizontal, as shown in Figure $\mathrm{P} 13.66 .$ At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring constant of 20.0 $\mathrm{N} / \mathrm{m} .$ Find the maximum distance the spring is compressed.

This problem covers the concept of the conservation of mechanical energy. Okay, And if you consider a our final position of the block, The gravitational potential Energy zero. Which means at this level, The gravitational potential energy is zero. Okay, so when the block is at the top, the spring is relaxed. Therefore The elastic potential energy of the spring is zero. The block is addressed. Therefore it's Connecticut disease, also syria. And when the block reaches to the uh to the base, the gravity is the potential energy will be zero. And when the block stops, it's kind of technology is also zero. Okay, now, that's right. Uh this in form of equation. So the initial gravitational potential energy is MG into H. And that equals the final electric potential energy. That is half of K. X squared. And from this we can right X. Equals two times MGH upon K and r squared. Now substitute the value to find the maximum compression in the spring. And that is equivalent spirit of To into the mass of the object is 500 g or 0.5 kg and two 9.8 m four seconds worth, Into the height edge, that is two m upon the spring constant 20 Newton for meat. So the maximum compression in the fingers 0.99 m

For this problem on the topic of momentum and collisions, we are told that two blocks A and B are forced together, which compresses a spring that is laying between them. The spring has a spring constant of 2500 newtons per meter, and the spring is compressed by three centimeters from its equilibrium length. We want to know the speeds of block A. And B after the blocks move off and the spring returns to its original length. Now they are essentially on a frictionless surface, and we'll use the conservation of momentum and the conservation of energy here to find the speed at which the blocks move off. Now the conservation of momentum tells us that the initial momentum of the system is equal to the final momentum, and the conservation of energy tells us that the initial energy is equal to the final energy. Now the initial momentum is zero since the two blocks are addressed and the initial kinetic energy E. I. Is equal to the energy stored in the spring, which is the elastic potential energy a half K data X squared. Now, since P I is equal to P F is equal to zero, we have M A V. A plus M B. The B is equal to zero, which means that M A V. A. The massive block a tendered speed is equal to minus the mass of block, B times its speed after it moves off. Now from the conservation of energy, the initial energy they put the final energy, which means that the initial elastic potential energy half K. Delta X squared. Where delta X. Is. The compression of the spring is converted all into kinetic energy. A half M. The A squared goes to block A plus a half M B. The b squared goes to block B. So simplifying this, we get okay, DELTA X squared is equal to M B squared over M A times V B squared less M B Vb squid. And so if we simplify further, we get okay. Delta X squared is equal to M B squared over M A plus MB all into V B squared, which means that we can rearrange and we can solve for Phoebe. And so we get Vb to be the square root of K. Delta X squared divided by M B into one plus M B of M A. And we can write an expression for V A. V A is simply minus M B of uh M A times B B. So we can calculate VB first by substituting the relevant values into that equation. And we get VB is equal to the square root of 2500 newtons commit to which is K. And the compression of the spring three centimeters or three times 10 to the minus two m squared, divided by three kg, which is the mass of B into one plus and be over M. A, which is three kg divided by the massive block A one kg. And so we get VB to be 0.43 the zero or 0.433 meters a second. Now that we have VB, we can find V. A. And V A. Is equal to minus and be three kgs over M. A, which is one kg times zero point 4330 meters per second, which gives us the speed of A. After the collision to be minus one 0.3 meters the second.

So here it's simply going to be the law of conservation of energy. Now, initially, we on Lee have gravitational potential energy. So we have m g h equaling, uh, the final mechanical energy, which in this case would only be potential energy of the spring. And so this would be equal in 1/2 times the spring, constant times the final compression of the spring squared. And so we can then say that for a incline h would be equaling. Uh, we can say x sign of Fada where this would, of course, be the angle of incline, and then acts would be the length of incline. So that's how we would find the height of the incline. And we can then say that M g h rather mg multiplied by x sign of fada going to be equal in 1/2 times k x final squared. And in this case, um, X final is not X. In fact, to avoid confusion, we should change this variable, and we can simply say no. So l would be the length of the incline. And at this point, we can solve for the compression, so X final would be equaling the square root of two em to M g l Sign of Fada, divided by K the spring constant. And so this is equaling the square root of two times the mass of 12.0 kilograms times 9.80 meters per second, squared times the length of 3.0 meters times sine of the angle of the incline, 35 degrees that's extend the square root. And this would all be divided by the spring constant of 3.0 times 10 to the fourth Newtons per meter. And we find that the final compression distance is equaling 0.116 meters. This would be our final answer for the compression of the of the final compression of the spring. That is the end of the solution. Thank you for watching.

Have a mess sliding down a frictionless incline uh coming in contact with the spring compressing the spring and then stopping. So in the beginning we can assume that the masses starting from rest, so the only energy that it would have ah would be the gravitational potential energy. And then at the end it comes in contact with the spring compresses the spring and then stops while the spring is compressed. So in this case we are taking the end of the problem to be the exact moment at which the spring is compressed and the spring is compressed and it causes the mass to come to a stop. So at that moment all of the gravitational potential energy has been uh converted into elastic potential energy. And specifically, we can say that then M. G. H. Some height associated with the incline would be equal to 1/2 times K. The spring constant for the spring multiplied by the final compression squared. And from the incline we know that then sine of theta. Sine of the angle of the incline would be equal to the height of the incline, divided by s the length of the incline, and so the height would be equal to s multiplied by sign of theta. And so we can then find the final compression. This would be equal to the square root of two mg. Multiplied by S. Multiplied by sine of theta, and this is then divided by the spring constant. Okay, we can then solve this would be equal to the square root of two, multiplied by the mass of 12 0.0 kg. multiplied by 9.8 m/s squared, multiplied by 3.00 m, multiplied by sign of 35°. Let's extend the square root. This is all divided then by the spring constant of 3.00 times 10 To the 4th. Newtons Per meter. And so the final Compression would be equal 2.116 m. This is the length, the length with which the spring is compressed. That is the end of the solution. Thank you for watching.