CATAGORY: Mechanics QUESTION: A 2kg block is released with an initial speed of 5 m/s along a horizontal tabletop toward a light spring with force constant 10 N/m. The coefficient of Kinetic friction between block and table is 0.25, and the block is initially 2.0 m from the spring. a) What is the max compression of spring? b) How far from the spring is the block when it comes to rest?

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The initial kinetic energy of the block is 1 1 (1) KE = - m v^2 = - 2kg (5m/s)^2 = 25 J 2 2 This energy is dissipated by friction AND goes into the potential energy of the spring. The potential energy of the spring is 1 1 (2) PE = - k x^2 = - 10 N/m x^2 = 5.0 N/m x^2 2 2 Where x is the compression length. the energy dissipated by friction is (3) EF = Uk mg (2m + x) = .25 2kg 9.8m/s^2(2m + x) where Uk is the coeficient of friction, m is the mass if the block and g = 9.8 m/s^2. We include a 2m + x because while the spring is compressing the block is still sliding. Most solution sets omit this part because x is usualy small compared to the sliding length but we include it out of formality. From conservation of all energy we have (4) KE = PE + EF Substitution of (1), (2), and (3) into (4) gives (5) 25 J = 5.0 N/m x^2 + 4.9(2m + x) or (6) 0 = 5.0 N/m x^2 + 4.9 N x + 9.8J - 25J 0 = 5.0 N/m x^2 + 4.9 N x - 15.2J or in terms of J and meters (7) 0 = 5.0 J/m^2 x^2 + 4.9 J/m x - 15.2J or dividing through by Joules (8) 0 = 5.0 (x/m)^2 + 4.9(x/m) -15.2 let y =x/m to give (9) 0 = 5.0 y^2 + 4.9y -15.2 Applying the quadratic formula to this gives ______________________ (10) -4.9 +/- \/ 4.9^2 - 4(5.0)(-15.2) y = ---------------------------------- 2(4.9) -4.9 +/- 18.11 = -------------- 9.8 = (-23.01/9.8 , 13.21/9.8) = ( -2.34, 1.35) x was taken to be positive so we have a comperssion of (11) x = 1.35m the initial potential energy of the compressed spring is then 1 (12) PE = - k x^2 = - 10N/m (1.35m)^2 = 9.1125 J ~ 9.11J 2 When the block reverses directio it slides back toward where it came from comming to a halt before its initial location because of friction. It covers the first 1.35m while the spring is still pushing and then a length L given by (13) 9.11J = Uk mg (L +1.35m) = .25 2kg 9.8m/s^2 (L +1.35m) = 4.9N L + 6.615 N m = 4.9N L + 6.615 J Thus (14) 2.495 J = 4.9N L L = 2.495 J/ 4.9N = .509 m ********************************************** * * * spring compression = 1.35m * * * * distance from spring when at rest = .509m * * * **********************************************