A spring of constant k = 10.2 N/m is attached to a wall. A 15.0 kg block is attached to the spring and allowed to oscillate on a frictionless surface as indicated in the figure. The mass of the spring is negligible. If the block is initially in the equilibrium position at time t = 0 with an initial velocity v = 2.00 m/s, what is the position at t = 0.70 s? Assume that the system is at the equilibrium position.

\fDe ( *= 0. 708 ) = (2.43 ) Sin Vm X 0- 70 = 2.43 Sin 10. 2 15 * 0.70 m = 2. 43 Sin ( 0-58 ) m X (+ = 0.73 ) = 0.62m so the position of block at t= 0.708 is, 0.02m