a) You jump off a bridge with a (massless) bungee cord (a stretchable spring) ti

Question

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) 15 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 (ie. this is how many g's it is exerting and we can tolerate about 4-5 g's before passing out. (1 mark) 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) 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)

Explanation / Answer

(A) Applying energy conservation,

PEi + KEi = PEf + KEf

{ PE = gravitational PE + rope potential energy}

(0 + 0) + 0 = (m g d - k (d-15)^2/2) + 0

60 x 9.8 x d = 55 (d - 15)^2 /2

588d = 27.5 d^2 - 825d + 6187.5

27.5 d^2 - 1413 d + 6187.5 = 0

d = 4.83 m or 46.5

d can not be smaller than 15m .

so d = 46.5 m

(B) Fnet = m a

k x - m g = m a

a = (55/60)(46.50- 15) - g

a = 19.08 m/s^2

a = (19.08/9.8)g Or 1.95 g

(C) 60 x 9.8 x d = k (d - 15)^2 /2

given d = 60 m

60x 9.8 x 60 = k (60 - 15)^2 / 2

k = 34.8 N/m

(D) Now 34.8 (60 - 15) - 60g = 60 a

a = 16.3

or a = 1.66 g

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