Question

A ride at an amusement park attaches people to a bungee cord, pulls themstraight down to the ground, and then releases them into the air. When theyare pulled to the ground, the bungee cord (which has a stiffness constant of35 N/m) is stretched 100 m beyond its unloaded length. What restrainingforce is required to hold a man with a mass of 80 kg to the ground just beforehe is released? (Recall that g = 9.8 m/s²)

Answer 1

Explanation

Step 1

given

$\begin{gathered} spring\text{ constant=35 }\frac{N}{m} \\ x=100\text{ m} \end{gathered}$

the force is given by:

$\begin{gathered} F=kx \\ replace \\ F=35\frac{N}{m}*100\text{ m} \\ F=3500\text{ N} \end{gathered}$

so, the force is 3500 N

Step 2

now,

Newton's first law states that if a body is at rest it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force, in other words the force due to the mass( weigth) + the restrainining force must be equal to the force of bungge, so the sum of forces equals zero

Free body diagram of the object

$\begin{gathered} Force\text{ of bungee-weigth-F}_{restrainn}=0 \\ \end{gathered}$

solve for F

$\begin{gathered} Force\text{ of bungee-weigth-F}_{restrainn}=0 \\ Force\text{ of bungee-weigth=F} \\ replace \\ 3500\text{ N-\lparen80kg*9.8}\frac{m}{s^2}\text{\rparen=F} \\ 3500N-784\text{ N}=F \\ 2716\text{ N=F} \end{gathered}$

therefore, the answer is

A. 2716 N

I hope this helps you