A 5.00-kg block is moving at ?v? = 6.00 m/s along a frictionless, horizontal surface 0? toward a spring with force constant ?k? = 500 N/m that is attached to a wall (?Fig. P6.79?). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of v ? 0?? (?Fig. P6.79?)

Solution 85P Here, the kinetic energy of the block is equal to the potential energy of the spring. Therefore, the necessary condition is, Kinetic energy of the block = Potential energy of the spring …..(1) Mass of the block m = 5.00 kg Speed v =06.00 m/s Force constant of the spring k = 500 N/m Let, the compression of the spring be x. From the condition given in equation 1, 1mv 2 = kx 2 2 0 2 mv 2 = kx 2 0 (a) Substituting the given values in this equation, 5.00 kg × (6.00) m /s = 500 N/m × x 2 2 2 x = 0.36 m x = 0.6 m Therefore, the spring will be compressed by 0.6 m. (b) Now, given x = 0.150 m 2 2 So, from the equation mv 0 = kx 2 2 2 5.00 kg × v 0 = 500 N/m × (0.150) m v 2 = 2.25 m /s 2 0 v = 1.50 m/s 0 The maximum value of v will 0e 1.50 m/s.