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Unformatted text preview: the compressed spring at A. Applying Equation 85, we find Wnc = Ilkmg = 6.K +!::l.U = U: = !kx2, since the kinetic energies at A and B and the change in gravitational potential energy are zero. Therefore, = ~kx2/J1.kmg = 0.5(340 N/m)(0.18 m)2/(O.27)x (1.5 kg)(9.8 m/i) = 1.39 m.
, FlGURE 837 Problem 52. hal and final positions of the block in "denoted by points A and B, (a) AKAB = 1)= ~(1.5 kg)(O (6.4 m/s)2) = 30.7 J. 2 *mg{YB  YA) = (1.5 kg)(9.8 m/s ) x ~()o= 25.0 J. (c) From Equation 85, AB+!::l.UAB =5.73 J. (d) The force of e incline is Ix = IlxN = IlKmg cos 30 direction of motion, so the work done ~sWi AB = ike = J1.Kmge cos 30. Thus; j,~QI./ngcos30 = 5.73 J/(1.5 kg)x " .:r4 m)cos30 = 0.13. Problem 54 Solution. Problem
55. A 2.5kg block strikes a horizontal spring at a speed of 1.8 mis, as shown in Fig. 838. The spring constant is 100 N/m. If the maximum compression of the spring is 21 em, what is the coefficient of friction between the block and the surface on which it is sliding? ','...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.
 Spring '08
 WORMER
 Physics, Conservation Of Energy, Energy, Work

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