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In one single stroke 2.0 . 103 kg weight free-falls 4.30 m before contacting steel rod and drives the rod 10.0 cm into the ground The weight remains in contact with...

Question

In one single stroke 2.0 . 103 kg weight free-falls 4.30 m before contacting steel rod and drives the rod 10.0 cm into the ground The weight remains in contact with the rod as they ' come to rest with no deformation of the rod or weight: (10) What is the total gravitational potential energy consumed in driving the rod? (10) What is the average force (F) exerted on the rod?

In one single stroke 2.0 . 103 kg weight free-falls 4.30 m before contacting steel rod and drives the rod 10.0 cm into the ground The weight remains in contact with the rod as they ' come to rest with no deformation of the rod or weight: (10) What is the total gravitational potential energy consumed in driving the rod? (10) What is the average force (F) exerted on the rod?



Answers

A Particle of mass $m_{1}=0.67 \mathrm{~kg}$ is a distance $d=23 \mathrm{~cm}$ from one end of a uniform rod with length $L=$ $3.0 \mathrm{~m}$ and mass $M=5.0 \mathrm{~kg}$. What is the magnitude of the gravitational force $F$ on the particle from the rod?

So the work that the beam does on the pile driver is we can say work and C N. C simply means non conservative. And this would be equaling f co sign of 180 degrees multiplied by Delta X. And so this would be equaling essentially negative F multiplied by 0.120 meters. Of course, this is Ah, this is co sign of 180 degrees because the angle between the force and the displacement vector are parallel, but the directions are opposite to one another. And so that's why we have co sign of 180 degrees now from the work energy through, um, we can say that the works of N. C is gonna be equaling. This would simply be equaling the kinetic energy final, minus the kinetic energy initial plus the potential energy final minus the potential energy initial. Now, this will essentially become 1/2 em multiplied by the final squared minus the initial squared plus M g multiplied by why final minus. Why initial? No, we can say that we're gonna choose why equal zero to be the level where the pile driver first contacts the top of the beam So essentially the driver starts from rest and comes to rest again. But why? Final? Equaling negative 0.120 meters and he's starting. Why initial being five positive. 5.0 meters. So here we have essentially negative F times 0.120 meters. Now here there. It's starting from rest and coming to arrest. So essentially we can eliminate these two and this entire term, then become zero and this would be equaling. Essentially an G or 2100 kilograms multiplied by 9.80 meters per second squared and this would be multiplied by negative 0.120 meters minus 5.0 meters. And we find that then the force here is equaling 8.78 times, 10 to the fifth Newtons and this is upwards. This would be our final answer. That is the end of the solution. Thank you for watching

Here's a diagram of the system. We cannot apply equation 13 1 directly. However, this is because the rod is an extended object. So essentially we have to consider a small differential element of the rod. We can say mass d m of thickness D. R at a distance are from M someone. So essentially we consider a differential mass. Sorry. The differential Force rather is going to be equal to the gravitational constant times amps of one times the differential mass divided by r squared. And so this is gonna be equal to G times M sub one multiplied by em over l times a differential thickness thes de are divided by r squared. And so at this point, though, we can just take the integral. So that's not too not too bad. Essentially, um because now we can just say that the force I was gonna be eat equal to the differential force and this would be equal to the gravitational constant times amps of one multiplied by, um, divided by l. And then we're gonna take the integral from D two l plus de of de are over r squared. Ah, we can say that f is not going to be equal to negative g m sub one mm divided by divided by AL multiplied by one plus out plus D minus one over D. And so this is gonna be equal to G m sub one m divided by D times L plus D, and we can then simply substitute so this would be equal to the force. So the force is then going to be equal to g times ems. Rather, we already have the equation here. So now we can just substitute the gravitational constant 6.67 times 10 to the negative 11th. This would be meters cubed per kilogram per second squared multiplied by 0.67 kilograms multiplied by 5.0 kilograms. And this would be divided by point to three meters times 3.0 meters plus point to three meters. And we find that the forces equaling 3.0 times 10 to the negative 10th Newton's. This would be our final answer. That is the end of the solution. Thank you for

For this problem, we will use the work energy to solve for the force being applied on the pile driver by the I beam. The work kinetic energy film says that work is equal to the change in kinetic energy. In this case, the power driver is initially at rest, which means that the initial velocity is zero and after the powder has hit the I. Beam and everything is settled, the power driver is also at rest, which means that the final velocity is also zero. This means that the kinetic energy, initial and final kinetic energy are both zero, which simplifies our equation to work is equal to zero. Now we have to think about what forces are being applied and what work they are doing. First we have the gravitational force, the work done by gravity. It's going to be equal to the mass times the acceleration due to gravity times the distance that the power driver travels. In this case we have the mass is 2100 kg and the distance it's going to be the initial five m that the pile driver drops, plus the 0.12 m that it drops after it has a the even which means that the work by gravity is equal to 2100 times 9.81 times 5.12, which is equal to 105,000 477 0.12 jewels. As you can see, the work done by gravity is positive because the gravitational force points downward and the pile driver is owned downward. Now we have to think about the force of the I beam applies to the power driver once it has hit the I. B, the work done by this force is going to be equal to the negative. The magnitude of force times the distance that is being applied. In this case, this work is negative because the power driver applies an upward the IBM applies an upward force on the pile driver and the pile driver is going downwards and it's going to be equal to negative the force which we are looking for. And the distance is this in this case is just 0.12 because this force is only being applied after the pot driver has hit the eye popping. Now we can plug these two works into our equation which is going to be the work done by gravity. Plus the work done by this force is going to be equal to zero, which gives us 105,000 477 0.12 minus 0.12 times the force is equal to zero. Now we can solve for the magnitude of the forest, which is going to be 878 Killer Newton's

So for party, we can first find the change in length. This would be equal to Alfa the linear expansion coefficient for steel multiplied by the original length times the change in the temperature. So this would be equaling 11. We're rather 11 times 10 to the negative sixth and this would be per Giger three Celsius so degree Celsius. The negative first multiplied by 2.0 meters, multiplied by the change in temperature of 40 degrees Celcius minus 20 degree Celsius. And so the change in the length is gonna be equaling 4.4 times 10 to the negative fourth meters. From this, the load exerts a force f equaling the weight. So for party, we can first find the change in length. This would be equal to Alfa the linear expansion coefficient for steel multiplied by the original length times the change in the temperature. So this would be equaling 11 or rather 11 times 10 to the negative sixth and this would be per degree. This would simply be 6000 kilograms multiplied by 9.8 meters per second squared and this is giving us 5.88 times 10 to the fourth Newtons. So we can then say that the work done on the rod would be equaling the force multiplied by the change in length or the allegation. So it should be 5.88 times the Celsius, so degree Celsius. The negative first multiplied by 2.0 meters, multiplied by the change in temperature of 40 degrees Celcius minus 20 degree Celsius. And so the change in the length is gonna be equaling 4.4 times, 10 to the negative fourth meters. From this, the load exerts a force f equaling the weight. This 10 to the fourth, Newton's multiplied by 4.4 times 10 to the negative fourth Newton's and this is giving us approximately 26 jewels of work being done on the rod. So this would be your answer for part a four part D. Then the energy added by heat Q E equals M C Delta T. This would be equaling 100 kilograms multiplied by this would simply be 6000 kilograms multiplied by 9.8 meters per second squared and this is giving us 5.88 times 10 to the fourth Newtons. So we can then say that the work done on the rod would be equaling the force multiplied by the change in length or the allegation. So it should be 5.88 times 488 Jules per kilogram, her degree Celsius multiplied by 20 degrees Celsius. And this is gonna give us 9.0 times 10 to the fifth Jules. And so from the first law of thermodynamics, the change in the internal energy would be equaling the heat transfer, plus the work done. And so this would be 9.0 times 10 to the fifth, plus 26 to 10 to the fourth. Newton's multiplied by 4.4 times 10 to the negative fourth Newton's, and this is giving us approximately 26 jewels of work being done on the rod. So this would be your answer for part a four part D. Then the energy added by heat Q E equals M C Delta T. This would be equaling 100 kilograms multiplied by fools, which is essentially just 9.0 times 10 to the fifth. Jules, essentially 26 jewels can be considered negligible when compared to 9.0 times 10 to the fifth. Jules. So this would be our answer for part C. That is the end of the solution. Thank you for watching.


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