Work Done by a Spring Force A perfectly elastic spring requires 0.54 J of work to stretch 6 cm from its equilibrium position (see figure (b). (a) What is its spring constant k? (b) How much work is required to stretch it an additional 6 cm? Strategy Work "required" means work done against the spring force, which is the negative of the work, that is W=??(z-22). For part (a), @A = 0 and 2B = 6cm; for part (b), @B = 6cm and 2 = 12cm. In part (a), the work is given and you can solve for the spring constant; in part (b), you can use the value of k, from part (a), to solve for the work. Solution 1. W = 0.54 J = {k[(6cm)2 - 0], sok = 3N/cm. 2.W ={(3N/cm) [(12 cm)2 - (6 cm)2] = 1.62 J. Significance Since the work done by a spring force is independent of the path, you only needed to calculate the difference in the quantity "kx2 at the end points. Notice that the work required to stretch the spring from 0 to 12 cm is four times that required to stretch it from 0 to 6 cm, because that work depends on the square of the amount of stretch from equilibrium, "kx:2. In this circumstance, the work to stretch the spring from 0 to 12 cm is also equal to the work for a composite path from 0 to 6 cm followed by an additional stretch from 6 cm to 12 cm. Therefore, 4W (0 cm to 6 cm) = W (0 cm to 6 cm) + W (6 cm to 12 cm), or W (6 cm to 12 cm) = 3W (0 cm to 6 cm), as we found above.
R6_6 1/1 point (graded) An object moves along a cubic path y = (0.25m-2) 23 from the origin A=(0,0) to the point B=(2m,2m) under the action of a force F = (5.0N/m) yi + (10N/m) x] . Calculate the work done. Give your answer in joules to three significant figures without units. 35 V 35 Submit Show answer R6_7 2/2 points (graded) The spring in the example above is compressed 6 cm from its equilibrium length. (a) Does the spring force do positive or negative work? positive work negative work it does no work. > (b) what is the magnitude of this work? Enter your answer in joules to two significant figures without units. 0.54 V