984 words - 4 pages

Lab: Finding the Spring Constant (k)I.Question/PurposeWhat is Hooke's Law, and what does it have to do with k?Objectives: Determine the spring constant of a spring, Calculate the elastic potential energy, Calculate gravitational potential energy, Determine whether mechanical energy is conserved in an oscillating spring.II.Materials ListMeterstick, Set of masses, and Support stand and clampIII.HypothesisIn order to estimate the spring constant (k), one must first understand that a higher value of the spring constant (k) represents a stronger spring, and a lower value represents a weaker spring. That is true because F and x in Hooke's Law (F=kx) have a direct relationship. So, I can state that k will be a numerical value and have the label of N/m. My guess for the spring constant (k) of the spring that we used is 10 N/m , because it seems to be a fairly strong spring, yet has some give.IV.Procedure: See AttatchedV.Recording and Analyzing DataAnalysis and Interpretation1.b) elongation = stretched spring - initial spring; initial spring = 0.00m;Therefore, elongation = stretched springc) F=mg; F=9.8mTrialgMass (kg)Force (N)19.80.10.9829.80.151.4739.80.21.9649.80.252.4559.80.32.9469.80.353.432.k = force/elongationTrialStretched Spring (m)Mass (kg)Force (N)k (N/m)10.040.10.9824.520.060.151.4725.330.080.21.9626.140.090.252.4526.950.110.32.9426.7260.130.353.4326.4average k = 25.99 N/m3.A stiffer spring would cause a higher value for the spring constant. It would increase the elastic and gravitational potential energies.4.b) elongation = highest point - initial spring; initial spring = 0;Therefore, elongation = highest point5.b) elongation = lowest point - initial spring; initial spring = 0;Therefore, elongation = lowest point6.x = highest pointk = 25.99PEelastic= 1/2kx21/2(25.99)x21/2(25.99)(.035)2 = .016 J1/2(25.99)(.054)2 = .038 J1/2(25.99)(.017)2 = .004 J1/2(25.99)(.075)2 = .073 J1/2(25.99)(.025)2 = .008 J1/2(25.99)(.046)2 = .027 J7.x = lowest pointk = 25.99PEelastic= 1/2kx21/2(25.99)(.065)2 = .055 J1/2(25.99)(.110)2 = .157 J1/2(25.99)(.022)2 = .006 J1/2(25.99)(.130)2 = .220 J1/2(25.99)(.045)2 = .026 J1/2(25.99)(.085)2 = .094 J8. The elastic potential energy is greatest when the spring is stretched the farthest (lowest point), such as in the case of Trial 4 (.13m), the elastic potential energy is .220 . The elastic potential energy is least when the spring is barely stretched (highest point), such as in the case of Trial 3 (.017m), the elastic potential energy is .004 . If this is thought of in terms of energy stored in the spring, the spring has more energy stored and ready to be released when it is stretched more. If the spring is not stretched, then there is no or little energy stored in the spring to be released.9. b) highest point = .17m, .17 - highest pointheight of mass = .17 - highest point.135 = .17 - .035.116 = .17 - .054.153 = .17 - .017.095 = .17 - .075.145 = .17 - .025.124 = .17 - .04610. b) lowest point = .17m, .17 - lowest pointheight of...

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