Can please solve question 4 and 5 explaining step by step on a picture please, thanks.all the information it's on the first picture. where Uk= 1/2K((x2)^2-(x1)^2)) and Ug= mg(x2-x1) and K is the slope and in this case is 7.99 N/m. I think this is enough information to solve this thank you.Just post the picute of all the step you did to solve it. m(kg)x (m)mg (N)k (N/m)100 0.106 m200kgl0.203m 1.960 7.99300kgb.352m9802940 N/M3.920400 hg 0421m500 kglosBm 4.91.960--980-7. 99N/•3X=4,9(1.002-.250)Slo pe•203-106mg-4.9%3DData Table 2Calculations Table 2X1mg(x2-X1)¼k(x;²-x1?)X2% Diff250m 1.002m% Diffs2.2% g-UR3.6848 3,763.185 3.25 2.0% gtUk2.71 2.76 1.8%.2.2542.36 4.8%.300ma5 m0.500 kg350 m,962m400mng s4.Data Table 3Calculations Table 3X1X2mg(x2-X1)V½k(x2²-x1?)% Diff7Som 40700m .4801.421 1.40 1,4%1.078 +10373.9%0.500 kg7203 +0.6776.2%650m . SOBo 600Le000%3.99 Sl.460)2-(.750)3) =-1.401 Follow-up Questions4. Instead of comparing changes in energy, compare the total energy at each point. Consider the firstcase of the 0.500 kg mass falling from x,=0.250 m to the value of x2 determined for that case. CalculateM 0.so the sum Uk + Ug at x1 as ½kx1² – mgX1. Now calculate that sum at x2 as 1/2kx22 – mgx2. Compare the twoand comment on the significance of the comparison.Som= lo5. Consider the same case as discussed in question 4. Calculate the value of x halfway between x1 and x2.Calculate Uk+Ug = ½kx? –mgx for that point. How does it compare with the values calculated inquestion 4? Explain the reason for this result.

In this question we have to compare the two quantities as asked in the question. Please give…

Follow-up Questions 4. Instead of comparing changes in energy, compare the total energy at each point. Consider the first case of the 0.500 kg mass falling from x1=0.250 m to the value of x2 determined for that case. Calculate 0.50 the sum Uk + Ug at X1 as ½kx1? – mgx1. Now calculate that sum at x2 as 1/2kx22 – mgx2. Compare the two and comment on the significance of the comparison. KEA PE

Follow-up Questions 4. Instead of comparing changes in energy, compare the total energy at each point. Consider the first case of the 0.500 kg mass falling from x1=0.250 m to the value of x2 determined for that case. Calculate 0.50 the sum Uk + Ug at X1 as ½kx1? – mgx1. Now calculate that sum at x2 as 1/2kx22 – mgx2. Compare the two and comment on the significance of the comparison. KEA PE

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter28: Special Relativity

Section: Chapter Questions

Problem 19CQ: The mass of the fuel in a nuclear reactor decreases by an observable amount as it puts out energy....

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Follow-up Questions 4. Instead of comparing changes in energy, compare the total energy at each point. Consider the first case of the O.500 kg mass falling from x,=0.250 m to the value of x2 determined for that case. Calculate the sum Uk + Ug at x1 as ½kx1? – mgx1. Now calculate that sum at x2 as 1/2kx2? – mgx2. Compare the two and comment on the significance of the comparison.
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