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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 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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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 m 200kgl0.203m 1.960 7.99 300kgb.352m 980 2940 N/M 3.920 400 hg 0421m 500 kglosBm 4.9 1.960--980-7. 99N/ •3 X= 4,9(1.002-.250) Slo pe •203-106 mg-4.9 %3D Data Table 2 Calculations Table 2 X1 mg(x2-X1) ¼k(x;²-x1?) X2 % Diff 250m 1.002m % Diffs 2.2% g-UR 3.6848 3,76 3.185 3.25 2.0% gtUk 2.71 2.76 1.8%. 2.2542.36 4.8% .300m a5 m 0.500 kg 350 m,962m 400m ng s4. Data Table 3 Calculations Table 3 X1 X2 mg(x2-X1) V½k(x2²-x1?) % Diff 7Som 40 700m .480 1.421 1.40 1,4% 1.078 +10373.9% 0.500 kg 7203 +0.6776.2% 650m . SOB o 600 Le00 0% 3.99 Sl.460)2-(.750)3) =-1.40 1
Transcribed Image Text:m(kg) x (m) mg (N) k (N/m) 100 0.106 m 200kgl0.203m 1.960 7.99 300kgb.352m 980 2940 N/M 3.920 400 hg 0421m 500 kglosBm 4.9 1.960--980-7. 99N/ •3 X= 4,9(1.002-.250) Slo pe •203-106 mg-4.9 %3D Data Table 2 Calculations Table 2 X1 mg(x2-X1) ¼k(x;²-x1?) X2 % Diff 250m 1.002m % Diffs 2.2% g-UR 3.6848 3,76 3.185 3.25 2.0% gtUk 2.71 2.76 1.8%. 2.2542.36 4.8% .300m a5 m 0.500 kg 350 m,962m 400m ng s4. Data Table 3 Calculations Table 3 X1 X2 mg(x2-X1) V½k(x2²-x1?) % Diff 7Som 40 700m .480 1.421 1.40 1,4% 1.078 +10373.9% 0.500 kg 7203 +0.6776.2% 650m . SOB o 600 Le00 0% 3.99 Sl.460)2-(.750)3) =-1.40 1
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 x,=0.250 m to the value of x2 determined for that case. Calculate M 0.so the sum Uk + Ug at x1 as ½kx1² – mgX1. Now calculate that sum at x2 as 1/2kx22 – mgx2. Compare the two and comment on the significance of the comparison. Som = lo 5. 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 in question 4? Explain the reason for this result.
Transcribed Image Text: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 x,=0.250 m to the value of x2 determined for that case. Calculate M 0.so the sum Uk + Ug at x1 as ½kx1² – mgX1. Now calculate that sum at x2 as 1/2kx22 – mgx2. Compare the two and comment on the significance of the comparison. Som = lo 5. 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 in question 4? Explain the reason for this result.
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