Question

te M of estion M A mass M = 1.2 kg is released from rest. From the initial height of 90 cm above the ground, the mass drops from an free spring (neither compressed, nor extended) of constant k = 580 N/m. The maximum compression of the spring, due to the impact of the dropping mass is dh = 180 mm. Analyse the system with respect to the balance of energy, and the apply the conservation of energy, considering the ground level as reference of the calculation of the potential energy and calculate: The energy at selected points indicated in the table: Potential Kinetic Elastic Ep1 = Eki = Polnt Eeli 1 J J J Epo = Eko = Eel point (Top of free o spring) J J J Ep2 = Ex21 = Eel2 = Polnt 2 J J J Calculate Vo = m/s the speed of the mass at the instant the mass hits the top of the free spring, the height of the free spring above the reference ho = (ground level) Two students are discussing about whether the acceleration is constant between points o and 2, and consider to calculate the speed at points 3 and 4 situated a respectively, Sh/2 from the top of the spring (point 3): V3 = m/s and (*) at point 4 which is situated 28h/3 below the top of the spring: : V4 = m/s and (*) The conclusion will be reached by applying vž – vu? = 2ax between any two pointed between o and 2 (*): Double point

Answer 1

point y Epuf Eeyt Eky a 10-5948 hy z horash 2618-16514- x 190 hua 498-16514 amm 251 3 Epya mghy a 1.2x9.81% 0.49816514 Epy = 5.86445 Een yak (sh - 1/24580 ( 22 +0.187² - 4.1465 Ely = 4.1763 Ek4 = 10.5948-5.8644-4.116 Ery 0.55445 muy? ossun 0.9612smis Just apply vaivo go aq ax (281 25h) 2 0.96195² - 9.3515 axa ( 2 x 0:18] 19.1898m 16² From BS acceleration is different it's clear that aceleration is not constant. so we cannot appy vo? Vigan formula between state o& state 2