Sol.

It will be assumed initially that the stiffness of the spring is small enough to allow the block to reach position B. The active-force diagram for the system composed of both block and cable is shown for a general position. The spring force 80x and the 300N. tension are the only forces external to this system which do work on the system. The force exerted on the block by the rail, the weight of the block and the reaction of the small pulley on the cable do no work on the system and are not included on the active-force diagram.

[ U = ]

U 1–2 = – = – 80.0 J

The work done on the system by the constant 300-N force in the cable is the force times the net horizontal movement of the cable over pulley C, which is – 0.9 = 0.6 m. Thus, the work done is 300 (0.6) = 180 J. We now apply the work-energy equation to be system and get

[U 1 – 2 = Δ T] – 80.0 + 180 = (50) (v 2 – 0)

v = 2.0 m/s Ans.

We take special note of the advantage to our choice of system. If the block alone had constituted the system, the horizontal component of the 300-N cable tension acting on the block would have to be integrated over the 1.2-m displacement. This step would require considerably more effort then was needed in the solution as presented. If there had been appreciable friction between the block and its guiding rail, we would have found it necessary to insolate the block alone in order to compute the variable normal force and, hence, the variable friction force. Integration of the friction force over the displacement would then be required to evaluate the negative work which it would do.