11:09 Search webassign.net Your last submission is used for your score. 1. [-/0.17 Points] DETAILS SCALCET9 6.4.AE.001. 0/100 Submissions Used MY NOTES ASK YOUR TEACHER Example 1 Video Example () (a) How much work is done in lifting a 1.3-kg book off the floor to put it on a desk that is 0.7 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s (b) How much work is done in lifting a 19-lb weight 5 ft off the ground? Solution (a) The force exerted is equal and opposite to that exerted by gravity, so the force is d+ 2 F = m - = mg = (1.3)(9.8) = and then the work done is W = Fd = ( (0.7) = (b) Here the force is given as F = 19 lb, so the work done is W = Fd = 19 . 5 = ft-lb. Notice that in part (b), unlike part (a), we did not have to multiply by g because we were given the weight (which is a force) and not the mass of the object Need Help? Read it 2. [-/0.17 Points] DETAILS SCALCET9 6.4.AE.004. 0/100 Submissions Used MY NOTES ASK YOUR TEACHER Example 4 Video Example () A 600-lb cable is 100 ft long and hangs vertically from the top of a tall building. (a) How much work is required to lift the cable to the top of the building? (b) How much work is required to pull up only 10 feet of the cable? Solution (a) Here we don't have a formula for the force function, but we can use an argument similar to the one that led to the definition of work. Let's place the origin at the top of the building and the x-axis pointing downward as in the following figure. 100+ We divide the cable into small parts with length Ax. If x,"is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x,". The cable weighs pounds per foot, so the weight of the i th part is 64x. Thus the work done on the ith part, in foot-pounds, is (64x) . X, = 6x, Ax force distance We get the total work done by adding all these approximations and letting the number of parts become large (so Ax - 0). W = lim 6x, Ax n -+ 20 100 100 6x dx = ft - lb (b) The work required to move the top 10 ft of cable to the top of the building is computed in the same manner as part (a). W 1 = 6x dx = 3x 2 0 - At - 1b Every part of the lower 90 ft of cable moves the same distance, namely 10 ft, so the work done is W2 = lim 10 64x 1=1 distance force ) 60 dx =[ ft - lb . (Alternatively, we can observe that the lower 90 ft of cable weighs 90 . 2 = 540 lb and moves uniformly 10 ft, so the work done is 540 . 10 = ft- lb. ) The total work done is W, + W2 = 300 + = 5,700 ft-lb. Need Help? Read It