A 55kg female bungee jumper fastens one end of the cord (made of elastic material) to her ankle and the other end to a bridge. Then she jumps off the bridge. As the cord is stretching it exerts an elastic force directed up on her. Calculate her acceleration at the instant the chord exerts an elastic force of 825N [up] on her.

Answers

Answer 1

Answer: Downward force = gorce of gravity (her weight) = 55 x 9.8 = 539 newtons / / / Upward force = 825 newton. / / / Net force = 825 - 539 = 286 newtons upward. / / / F = m a / / / a = f / m = 286 / 55 = 5.2 meters per second squared Upward. / / / Note: We don't know the length of her fall before the bungee kicks in. At the moment we're dealing with, when her acceleration is 5.2 m/s-squared Upward, her velocity can very well be Downward.

Answer 2

Answer: When the jumper jumps off , there is always a force of gravity acting on her which is 55*9.8 =539N downwards. (Considering the acceleration due to gravity to be 9.8)
Since the question says that the force exerted by cord is 825N upwards , it is obvious that the pseudo force of acceleration must act downwards and the jumper must be moving upwards.
Assume the acceleration at that instant be 'a' .
Pseudo force = mass * acceleration = 55*a
Since the body is in dynamic equilibrium , all forces must cancel out.
Hence ,
539 +55*a = 825
=> 55*a = 825-539 = 286
=> a = 286/55 = 5.2
So , acceleration at that instant was 5.2 m/s²

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Answers

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