Question

A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant of 35 N/m. What is the greatest vertical distance below the bridge surface that the bungee jumper will reach? Enter your answer as a positive number in meters, but do not enter units.

Answer 1

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

Answer 2

Answer:

$X=74.7$

Explanation:

From the question we are told that:

Mass $m=100kg$

Length $l=10m$

Spring constant $\mu=35N/m$

Generally the equation for potential energy of mass is mathematically given by

 $P.E_m=mgh$

Since

 $P.E_m=P.E_s$

Where

 P.E_s =potential energy of spring

Therefore

$m*g*(x+10) = 0.5*k*\mu^2$

$100*9.8*(x+10) = 0.5*35*\mu^2$

$980*(x+10) = 17.5*\mu^2$

$980*x+9800 = 17.5*\mu^2$

 $17.5*\mu^2 - 980*\mu - 9800 = 0$

Comparing the equation above with standard quadratic equation

 $17.5*\mu^2 - 980*\mu - 9800 = 0$

 $ax^2+bx+c=0$

Giving

  $a=17.5\\ b=-980\\ c=-9800$

Solving Quadratic equation the roots of the equation is given as

 $\mu_1=64.66$

 $\mu_2=-8.661$

Since

$\mu$ can not be -ve

Therefore

The vertical distance attained by the bungee jumper is given as

 $X=\mu+l$

 $X=64.7+10$

 $X=74.7$