Physics 101 help majorly needed?

a) calculate the stop distance d below the bridge

gravity potential energy will convert to spring potential

½kx² = mgh

if d is the total distance of drop h, d - 15 is the stretch of the cord x.

½k(d - 15)² = mgd

27.5(d² - 30d + 225) = mgd

27.5d² - 27.5(30)d + 27.5(225) = 60(9.8)d

27.5d² - 825d + 6187.5 = 588d

27.5d² - 1413d + 6187.5 = 0

quadratic formula

d = (1413 ±√(1413² - 4(27.5(6187.5))) / (2(27.5))

d = 4.83 m which we ignore as the spring has not yet begun to stretch

or

d = 46.548

d = 47 m ANSWER

to the 2 sig. fig. of the question numerals

b) What is the magnitude of the acceleration the cord exerts on you when it is fully extended?

Force in the spring will equal mass times acceleration

F = kx = ma

55(46.548 - 15) = 60a

a = 28.919 m/s² / 9.8 m/s²/g

a = 2.9509...

a = 3.0 g's

c) What spring constant, k, for full dunk 60 m below?

½kx² = mgh

½k(60 - 15)² = 60(9.8)60

k = 34.844...

k = 35 N/m

d) Maximum upward acceleration of (c)

34.844(45) = 60a

a = 26.133 m/s² / 9.8 m/s²/g

a = 2.6666...

a = 2.7 g's

As this is a lower acceleration rate, it is the "softer" ride.

I hope this helps.

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a) You jump off a bridge with a (massless) bungee cord (a stretchable spring) tied around your ankle. You fall for 15 m before the bungee cord begins to stretch. Your mass is 60 kg and we assume the cord obeys Hooke’s law, F = –kx, with k = 55 N/m. If we neglect air resistance, use conservation of energy to calculate the distance d below the bridge your foot will be before coming to a stop. (3 marks)

g is approximated to 10 m/sec^2

m*g*(h+x) = k/2*x^2

60*10*(15+x) = 27.5*x^2

27.5x^2-600x-9000 = 0

x = (600+√600^2+110*9000)/55 = 32.0 m

H = h+x = 15+32 = 47.0 m

b) What is the magnitude of the acceleration the cord exerts on you when it is fully extended? Give answer in terms of multiples of g = 9.8 m/s2 (i.e. this is how many g’s it is exerting and we can tolerate about 4-5 g’s before passing out). (1 mark)

F = k*x = 55*32 = 1760 N

top deceleration a max = (1760-60*g)/m = (1760-600)/60 = 1160/60 = 19.3 m/sec^2 ≈ 2g negatives

c) If the river is 60 m below the bridge, what spring constant, k, do you need for the bungee cord so that you get your body fully dunked in the river (i.e. your foot is at the level of the river, 60 m below the bridge)? (2 marks)

m*g*(15+45) = k/2*45^2

2*60*10*(15+45) = k*45^2

60^2*20 = k*45^2

K = 60^2*20/45^2 ≈ 35 N/m

d)What is the magnitude of the acceleration the cord exerts on you when it is fully extended for the spring constant found in (c)? Again, give answer in terms of multiples of g = 9.8 m/s2. Which is the softer ride? the bungee cord used in (a) or c)? (1 mark)

F = k*x = 35*45 = 1575 N

top deceleration a max = (1575-60*g)/m = (1575-600)/60 = 975/60 = 16.2 m/sec^2 ≈ 1.6 g negatives

softer ride is that with k = 35 N/m