Question

A bungee jumper has a mass of 60kg and uses a 25m long bungee cord (unstretched length) with an elastic coefficient of 800N/m. a) What is the maximum speed of the jumper? b) What is the maximum amount of stretch that will occur when the jumper drops from rest from a very high bridge?

Answer 1

Answer:

a ) 2.68 m / s

b )  1.47 m

Explanation:

The jumper will go down with acceleration as long as net force on it becomes zero . Net force of (mg - kx ) will act on it  where kx is the restoring force acting in upward direction.

At the time of equilibrium

mg - kx = 0

x = mg / k

= (60 x 9.8 ) / 800

= 0.735 m

At this moment , let its velocity be equal to V

Applying conservation of energy

kinetic energy of jumper + elastic energy of cord = loss of potential energy of the jumper

1/2 m V² + 1/2 k x² = mg x

.5 x 60 x V² + .5 x 800 x .735 x .735 = 60 x 9.8 x .735

30 V² + 216.09 = 432.18

V = 2.68 m / s

b ) At lowest point , kinetic energy is zero and loss of potential energy will be equal to stored elastic energy.

1/2 k x² = mgx

x = 2 m g / k

= (2 x 60 x 9.8) / 800

= 1.47 m

Answer 2

Answer:

a)

maximum speed of jumper is 22.14m/s

b)

Maximum stretch that will occur,x=94.3 m

Explanation:

Answer:

Explanation:

In this question we have given

mass of bungee jumper,m=60kg

length of cord,h=25m

elastic coefficient, k=800N/m

We have to find the maximum speed of the jumper

here law of conservation will be applied it means the potential energy lost by bungee jumper when he reaches at a specific depth is equal to the sum of strain energy which is produced due to stretching of bungee cord and kinetic energy gained.

It means

U=KE+SE....................(a)

Here potential energy of bungee jumper,U=mgh..................(1)

Put values of m,g and h in equation 1

we got,

$U=60kg\times 9.8ms^{-2}\times25m$

U=14700 J...........(2)

kinetic energy is given as

KE=$\frac{mv^2}{2}$..............(3)

put value of m in equation 3

KE=$\frac{60v^2}{2}$

$KE=30v^2$.............(4)

Strain energy is given as

SE=0 (when bungee cord is unscratched)

put values of KE and U in equation a we got

$30v^2=14700\\v=22.14 ms^{-1}$

maximum speed of jumper is 22.14m/s

b)

maximum amount of stretch be x

than SE=k(x-25)

put valuse of U,KE and SE in equation a

$60\times 9.8\times (x+25)=\frac{30\times 22.14^2}{2}+800(x-25)\\588x+14700=14700+800x-20000\\212x=20000\\\\x=94.3 m$

Maximum stretch that will occur,x=94.3 m