Principle of Work and Energy for a System of Particles 6 of 20 To apply the prin

Question

Principle of Work and Energy for a System of Particles 6 of 20 To apply the principle of work and energy to a system of The principle of work and energy can be extended trom one partide to include a system of particles as folows As shown, a 260 lb bungee jumper wants to jump from the top of a 455 ft high bridge and be able to just souch the water belowFigure 1) Given that the bungee cord has a spring constant of 576 Ib/ft what should the bungee cord's relaxed length, I, be for this jump? Express your answer numerically to three significant figures and include the appropriate unts Ti is the system's ritial knete energy View Avalilable Hints) U 3 is the sum of the work done by all external and ntemal forces acting on the system's particles, and T, isto system's firat knetic oney wher,tho particies ether are undergoing only translational motion or are connected by inextensible cablesthe forces of the particles negate each other, only the external forces are required in the priniple of work and Value Units When fretional forces are present, the work done by the fricionl force is UypNs, where y is the oeficent of kinetic friction, N is the normal fonce, and s the body's dspiacement in this equation, the work done by riction rresents both the esternal work of friction and the inenal work that is cenverted i Part B As shown boas resing on anorert ncines, connected by an niatic cable that passes over a motor ess pley(ero 2). Block A weighs 11 0 lb and block B weighs 48.0 1b. The indine angles are 0 28 0 degrees and 45.0 degrees, and the coefficient of sinolic tion betesen the biocks and he inclines i0.150. When the bdocks are rcleased trom rest, bock Bsides down ts incine. What is the magnitude of block A's velocity, A,ahor block 8 has slid 1.30 ft7 Figure 102> Express your answer to three significant figures and include the appropriate units View Avaliable H AValue Units Neat >

Explanation / Answer

If jumper just touches the water that means speed at 455 ft below will be zero and initially speed was also zero so there is no change in KE.

as it goes down his potential energy increases that must be equal to energy stored in cord.

so mgh = kx^2 /2

260*455 = 5.76x^2 /2

x = 202.67 ft

x = stretched length - relaxed length

relaxed length = 455 - 202.67 = 252.33 ft

(2nd Part)

(48x1.30 sin45 - 11x1.30 sin28) - (0.150x48x1.30xcos45 + 0.150x11x1.30 cos28) = ((48 + 11)/32.17)v^2 /2

28.90 = 0.917 v^2

v = 5.61 ft/s

Please rate my answer, good luck...

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