# Raj speed of 64 ms it comes to a halt afteri solution

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Unformatted text preview: the compressed spring at A. Applying Equation 8-5, 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 8-37 Problem 52. hal and final positions of the block in &quot;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 8-5, 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 &quot; .:r4 m)cos30 = 0.13. Problem 54 Solution. Problem 55. A 2.5-kg block strikes a horizontal spring at a speed of 1.8 mis, as shown in Fig. 8-38. 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.

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