Question

The system is released from rest with the spring initially stretched 5.8 in. Calculate the velocity v of the cylinder after it has dropped 1.5 in. The spring has a stiffness of 3.5 lb/in. Neglect the mass of the small pulley.

Chapter 3, Problem 3/142 (video solution to similar problem attached) XIncorrect The system is released from rest with the spring initially stretched 5.8 in. Calculate the velocity v of the cylinder after it has dropped 1.5 in. The spring has a stiffness of 3.5 lb/in. Neglect the mass of the small pulley k= 3.5 lb/in. 96 lb 3514 Answer: ft/sec the tolerance is +/-2%

Answer 1

When the block is at rest, the initial energy is equal to the potential energy of the spring that 1s, When the block drops by 1.5 inc., the spring elongates by. 2(1.5 inc.)-3.0inc. Therefore, the final energy is as follows (3.5 lb/in.) (5.8 in35 lb/in.)(58 in. +3.0 inc.)-(96 1b)g(1.5 inc(961b)v Solve for v (3.5 Ib/in.) (5.8 in.) (3.5 lb/in.) (5.8 in. +3.0 inc.)-961b) (386) (1.5 in.)+(96 Ib)v 58.87 lb in.-135.52 1b-in. -55,584 lb in. +48 lb v* Solve for v 58.87 1b in.-135.52 lb in.-55,584 1b in.+481b v v 34.0 m/s