### Spring Constant and Hooke's Law Assignment

Added on - 21 Apr 2020

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1Theoretical questiona)Spring constantF= KETaking a single point from the line graph, say (20,56), where we have a force of 20N andextension of 58 mm.Converting the extension into meter = 56/1000 = 0.056 mSpring constant (K) = force/extensionK= 20/0.056K = 357.14 N/Mb)Energy stores in the spring at any certain point of force application = PE = KE2The application of 25N coincides with an extension of 70mm (0.07m)Therefore energy at this point will be;PE = ½*357.24*0.072= 0.875 Jc)Force required to extend the spring 300cm (3m)F =KEF= 357.14 * 3= 1071.42 Nd) Assumptions made at c) aboveThe spring extension limit is not achieved and that the spring extends uniformlyThe force and extension are one dimensional vector meaning they are acting in the samedirectionThe Hookes law is obeyed2)Experiment task

a) Table of resultsActual mass (g)Mass (g)Extension (cm)150.25501.9250.191003.9350.141505.8450.372007.6550.182509.5b) Force vs extension graphAssuming the gravitational acceleration to be 9.8m/s2Mass (g)Force (N)Extension (cm)Extension (m)500.491.90.0191000.983.90.0391501.475.80.0582001.967.60.0762502.459.50.0950.0190.0390.0580.0760.09500.511.522.53Force (N) Vs Extension (M) GraphForce(N)c) spring constant and elastic potential energythe spring constant may be found through calculation of the gradient of the line graphtaking two point as (0.019,0.49) and (0.095,2.45)the gradient will be2.45−0.490.095−0.019

= 1.96/0.076 = 25.79 N/mPE = ½*K*extionsion2PE = ½*25.79*0.0952= 0.116375 Jd)The spring is a Hookean. From the graph plotted, it is clear that the spring is able to obey theHookes law. The graph is a straight line showing that force is directly related to extension.According to Hookes law, F =KE, where K is a constant known as the spring constant. Thisspring is able to obey this rule and therefore can be regarded as a Hookean.3)Theoretical questiona) Mass and weight of the beamAssuming that the weight of the beam is x N/mTaking the moments about the pivot to create the equilibrium situation(1.5*0.15)+ (x *0.3*0.3/2) = (x*0.4*0.4/2) – (0.4*1.5)0.225 + 0.045x = 0.08x – 0.6Collecting like terms together0.035x = 0.825X = 23.57 N/mThe weight of the beam will be 23.57 N/m * 0.7 m = 16.5NAssuming gravitational acceleration to be 10m/s2Mass = 16.5/10 = 1.65 kgb) Assumption on part a)the assumption is that the beam has weight which affects the equilibrium statec) Reaction at pivotThe reaction at the pivot will be

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