Question

A block with mass m -6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v 4.9 m/s. The block oscillates on the spring without friction "What is the spring constant of the spring? N/m You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question, 2. What is the oscillation frequency? Hz You currently have 1 submissions for this question. Only 10 submission are allowed You can make 9 more submissions for this question "After t - 0.29 s what is the speed of the block? 292 You currently have 5 submissions for this question. Only 10 submission are allowed. You can make 5 more submissions for this question. "What is the magnitude of the maximum acceleration of the block? 33.32 m/s You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. "At t -0.29 s what is the magnitude of the net force on the block? You currently have submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. "Where is the potential energy of the system the greatest? At the highest point of the oscillation At the new equilibrium position of the oscillation, At the lowest point of the oscillation. You currently have 6 submissions for this question. Only 10 submission are allowed. You can make 4 more submissions for this question.

Answer 1

Solution 1 As the weight is hung the to a distance n-0.22m equilibrium. So, spiry stretches and attains static 1 Efy=0 wuuu Goxmg - kn =0 D a 6.8 x 9.8 - K x 0.22 - 0 7303 Am k = 30. 29 Nlm Any ft şi 0.34 Hertz Ans If = We know - 3 amplitude Ea) = 0.22m, w = 27f= 0.681 t = 0. 295 U= awcoswt = 0.22X 0.888I cos (6.68TX 0.29) Lu= 0.38 m/s

4) Maximum acceleration = w²a = (276) 2 x 0.22 = 2.95 mls?