1-As shown, a 160 lb bungee jumper wants to jump from the top of a 220 ft high b

Question

1-As shown, a 160 lb bungee jumper wants to jump from the top of a 220 ft high bridge and be able to just touch the water below(Figure 1) . Given that the bungee cord has a spring constant of 8.63 lb/ft , what should the bungee cord's relaxed length, l, be for this jump? 2-As shown, two blocks, resting on different inclines, are connected by an inelastic cable that passes over a frictionless pulley(Figure 2) . Block A weighs 17.0 lb and block B weighs 50.0 lb . The incline angles are 1 = 30.0 degrees and 2 = 51.0 degrees , and the coefficient of kinetic friction between the blocks and the inclines is k = 0.300. When the blocks are released from rest, block B slides down its incline. What is the magnitude of block A's velocity, vA, after block B has slid 2.50 ft ?

Explanation / Answer

Given,

Weight of bungee jumper, W = 160 lb

Height, H = 220 ft

Energy to be absorbed by cord = Potential energy of jumper = 220 x 160 = 35200 ft.lb.

Let extension of cord be y, then maximum force in cord = 8.63 y

Energy absorbed by cord = 1/2 x 8.63 y x y = 4.315 y2 = 35200

Therefore, y = sqrt(8157.59) = 90.32 ft

Since the total distance to fall is 220 ft and the extension is to be 90.32 ft, the free fall before extension starts = relaxed length = 220 - 90.32 = 129.68 ft.

according to chegg guidelines we can solve one problem at a time so please re-attach your other question seperately..thank you goodluck

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