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534 block is released from rest at height ho bove vertical spring with spring constant 290 N/m and negligible mass; The block sticks to the spring and momentarily s...

Question

534 block is released from rest at height ho bove vertical spring with spring constant 290 N/m and negligible mass; The block sticks to the spring and momentarily stops after compressing the spring 12.5 cm_ How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of ho? (d) If the block were released from height Sho above the spring what would be the maximum compression of the spring?(a) NumberUnits(b) NumberUnits(c) NumberUnits(d) NumberUnit

534 block is released from rest at height ho bove vertical spring with spring constant 290 N/m and negligible mass; The block sticks to the spring and momentarily stops after compressing the spring 12.5 cm_ How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of ho? (d) If the block were released from height Sho above the spring what would be the maximum compression of the spring? (a) Number Units (b) Number Units (c) Number Units (d) Number Units



Answers

A 700 g block is released from rest at height $h_{0}$ above a ver-
tical spring with spring constant $k=400 \mathrm{N} / \mathrm{m}$ and negligible mass.
The block sticks to the spring and momentarily stops after compressing the spring 19.0 $\mathrm{cm} .$ How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of $h_{0} ?$ (d) If the block were released from height 2.00$h_{0}$
above the spring, what would be the maximum compression of the
spring?

For this problem on the topic of conservation of energy, We are told that a block with mass 700 g is released from a height h not above a vertical spring. The spring has a spring constant of 400 newtons per meter, and the bloc lands on the spring sticks to it and momentarily stops while compressing the spring by distance of 19 centimeters. You want to know the work done by the block on the spring. The work done by the spring on the block. The value for H. Not and you want to know the maximum compression of the spring if the block were released from a height of two H. Not. Now, firstly we have if we place the reference position for evaluating gravitational potential energy at the relaxed position of the spring, and we use X. For the springs compression measured positively downward. We have when X is zero 0.19 meters. We have the work W. S. To be minus a half. Okay, X squared. And this is the work done by the block on the spring and this is minus seven 0.2 jules. Now we know that by Newton's third law, the work done by the block on the spring is the same as the work done by the spring on the block. So this work is seven 0.2 jules for part C. We want to find the value of H. Not now we use energy conservation. We have the initial kinetic energy plus the initial potential energy of the system must equal the final kinetic energy plus the final potential energy. So initially the kinetic energy of the block is zero since it starts arrest. And it has a gravitational potential energy a half. M. G H. Not the final kinetic energy zero since the the block momentarily comes to rest. And the final potential energy is that due to elastic potential energy, which is a half K. X squared minus the gravitational potential energy MG. X. Now with m equal to 0.7 kg, we rearrange this equation we get H note to be zero 0.86 m. Lastly for part D. Of the problem, we have a new value for the height. We call it H not crime and this is equal to two H. Not. And from before this we can see that this is 1.72 m. Now we solve for a new value of X. Using the quadratic formula and we have mg. H. Not prime equal to minus MG. X. Plus a half. Okay, X squared. And so by setting up the quadratic formula, we have X. To be MG plus the square root. Uh huh. Mg squared plus two mg. Okay. H not prime divided by the spring constant K. If we substitute our values and we get the new compression of the spring to be 0.26 m.

All right. So in this problem, we've got a mass that's being dropped onto a spring. It's going to compress the spring. Um, some distance in this case the distance is 19 cm or a 0.19 m. The mass of the block is .7 kg or 700 g. We're told what the spring constant is, it's 400 newtons per meter. And we need to figure out a couple of things. So the first part part A we need to figure out how much work was done on the spring. Yeah, how much work the block did on the spring and then part B. We'll figure out how much work the spring did block. So to figure out the amount of work the blocked it on the spring, we can just look at our change in potential energy is to work where this is going to be the elastic potential energy of the spring. So the work done on the spring is going to be Uh negative one half. Okay, X squared rex is going to be a negative 0.19 m because we're pushing it down. So if we plug in our values here we get 1/2 times 400 times .19 squared. And We'll get the value of 7.22 jewels or negative 7.22 jewels. And it's negative because it's in the downward direction in this case. Mhm. And then in part B how much work did the spring due on the block? Well, it'll do the exact opposite. So conservation of energy in this spring block system, If the block did 7.22 jewels on the spring, the spring did 7.22 jewels on the block. But in this case a positive energy to make sure that her total energy stayed the same between the two of them. Now, for part C, we need to figure out what this height is. So this is a church one. We need to figure out how high it was dropped. Okay, so we've got the potential energy of the mass is going to start way up here. And so we're going to have the gravitational potential energy of the mess from this height. And also a gravitational potential energy from that little spot there. And that has to equal be changing potential energy due to the elastic spring. So we're gonna have MG. H. One plus MG. X. Where X is this compress in the amount that the spring was compressed, compression hate. And then that's going to equal that one half K. X. Squared. And so if we solve this for H1, we're gonna get K. X squared over two mg minus X. So if we divide both sides by MG, we're gonna get K. X squared over MG times one half or two MG. These MGs go away and we've got an X. Over here. So we just subtract X from both sides. We get each one do this amount right here. If we plug in our numbers we're gonna get a height of 86.2 0.8 62 meters As our each one. Now the last part of this problem is to figure out how far the spring would compress if we doubled that height if we went to H. Instead. And so for this we're going to use this exact same equation right here to solve this. So now instead of M. G. H. One it's going to be too MG. Each one this is going to be a new X. So we'll call this next one and that's gonna one half K. X. One square. And so we're gonna be solving for this next month. So we just, the only thing we did was we put that to here and so we can solve this. This is just going to be a quadratic equation. So I just need to be order everything. So we get this is our a minus energy. X one minus two. M. G. H. one. So this is A. B. C. And we can use our quadratic Formula two. So this so excellent is going to be equal to um MG plus or minus square root squared minus four times and a half. Okay, times negative. Two G. And so under the square root Divided by two times 1 half. Okay, if you plug those numbers in for everything, you'll get a value. Next one being 0.262 m or 26.2 centimetres.

All right. So, uh, this problem So we have a spring but touched on the ground. And then we have a book. And when When the book is ah. Reached. When? When the blood reached the spring. They're attached. Okay. Ah. Okay, So the mess of the book is ah em equal to 0.25 kilograms and a spring constant k. Be quote to ah, 2.5 new to the centimeter, or it's your 50. Newt says what made a okay. And ah, so we know that when the spray So that makes the maximum comprende? The maximum compressed the distance of the spring. It's ah h equal 0.12 meters. So, uh, part, eh? I want to find out the gravitational force, the work done by the gravitational force in this process. So the work done by the gravitational force is simply mth Okay, so there's a m and G is like a gravitational constant. And here's H is gonna find out. The video is zero points. 29 Jos. Mr. Jos. Okay. And for part B, uh, we want to find out that the work done by the spring falls so ah, one the When the block reaches the lowest of the low is the place right here. So the work done by the spring ws we'll be good to have okay times h swit. But in this case, this is a negative because they're great because the forces ah is opposite to the displacement off the block. So there's a Maybe So it s okay. Ihsaa case here to 50. And H is here so you can find out the Valerie's negative 1.8 jos the night you compare this to results. You see that the work done by the springs actually smaller. I mean, the magnitude of this work is actually larger than the magnetar off the gravitational force. So the rest of you comes from the initial kind of the energy of this book. So, uh, to find out that this the initial kinetic energy we can we can use Ah Hee hee for you. Okay. The kinetic energy initial kinetic energy plus W g, which is the work done by the gravitational force equal to the work. So now they're not able to in a plus. Ah, sorry. And pause. The work done by the spring equal to zero so we can sew for the UK equal 1.51 Jos. So we're going to solve for the initial velocity B I equal to three 0.47 member second, okay, And for party. So say that if the speed at impact is doubled, what is the maximum compression of the spring? So let's say B I in this case April 22 b I So this is a new initial velocity which is doubled so equal to 66.94 made up a second. So again, we can utilize this relation. Uh, so w s equal negative UK minus w g and the Because we have a new compression at this time. So w s is a prime is the prime. This is also primed. We have Ah, negative, huh? Okay. H prime squid equal to Ah, negative. Uh huh. And the I primed squid miners this age prompt minus mg age prime. Okay, so in this equation, we can see that that we already know. Okay, we know am And we nog. So the only on little variable is this h prime you can solve. You can sell for this age Prime by by just self this equation. Ah, so h prime. You will see that equal to zero point 23 meters, okay?

Mhm. In the party of this problem, we are going to calculate the verden by the gravitational forces. That is W G. The president should be equals to the again in potential energy. So we can right here this WGZ equals two gaining potential energy. And we can watch this potential energy as one divide by two. Okay square. In this case the supreme constantly. And is the amplitude of the oscillations. So this is the last two potential energy That set the values into the square. So here we have WG is equals to one, divide by two into Spring constant, which is equals to 2.5 multiple laboratories power to Eur 10 per meter into 12. multiply with Tony's power minus two m old square. So from here we will get the value for this. W G S W Z equals two. 1.2. Here, joe. This is the answer to the first part of this problem. Now, let's move to the second part of this problem. In this case we are going to calculate the verdant by the spring force that is denoted by W. Right? Yeah. Since the screen force is opposite to the displacement, so the virgin will be negative for this case. So we can right here that this would then will be negative of W. G. And This is equal to -1.9 in the party of this problem, we have to find the speed of the blog just before it hits the spring so that the speed of the blockade. This cases we we know that the kind of energy will be equals to the potentially there's a story in the spring. So we can right here. The kinetic in education should be equals two potential energy stored in the spring. And we can wipe this uh kinetic energy has one divide by two and with square has the same as the mass of the block. And visit speed Should be equals two. one divide by two K. A. square. And this recalls to the W. So here we can wipe this question for the v. As well as equals two to W. Divide by em. And we have square on it squirreled on it. Okay, so let's set the values into the square, Squirreled off two into 1.8 years old Divide by this mess which is equals to 200 and 50. multiply with journalist borrow -3. Hello Graham. So from here we will get the value for this. We as well as equals two three pointed metal prosecuted. Let's move to the party deal with this problem. So in this case we are going to find the compression in the spring of the speed of the impact is uh the belt speed speed of the blockade. The impact is the belt. So this time we need to calculate the expression compression in the spring said this is a pride if the speed of the block and the impact is the built. So let's set equation by applying the law of conservation of energy That is written as one, divide by two K. A Prime Square is equals two. When you add it by two I am. We prime square. So here this week prime is done in Sweden and the prime is corresponding of compression in the spring. So from here we can avoid the screen for a prime as a primary equals to square it off and divided by K. We have pride. From here we can say that this uh prime is proportional to re prime. This proportionality says that if the speed is doubled, the compression will be doubled because these two are linear similarly depending on each other. So say if if this week is doubled, like if we Prime is equals to two v. Then it that case this air prime will be equals to two weeks to A. So we have a prime is a prime is equals to two in two, Which is equal to 12 cm. So the new compression will be equals to 24 cm. Thank you.


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