Question

A ride at an amusement park attaches people to a bungee cord, pulls them

straight down to the ground, and then releases them into the air. When they
are pulled to the ground, the bungee cord (which has a stiffness constant of
35 N/m) is stretched 100 m beyond its unloaded length. What restraining
force is required to hold a man with a mass of 100 kg to the ground just
before he is released? (Recall that g = 9.8 m/s²)

Answer 1

`2520\; {\rm N}` downwards (assuming that the weight of the bungee cord is negligible.)

Step-by-step explanation:

To find the magnitude of the unknown force, start by considering all the forces acting on this person. Assuming that this person isn't moving, derive an equation from the fact that the resultant force on this person should be zero. Solve this equation to obtain the magnitude and orientation (upward or downward) of the unknown force.

Under the assumptions, forces on this person include:

• Restoring force from the bungee cord (upward,)
• Weight of the person (downward,) and
• Unknown force on the person.

With a spring constant of
`k = 35\; {\rm N\cdot m^(-1)}` at
`x = 100\; {\rm m}` from the equilibrium position, the restoring force from the bungee cord would be:

`\begin{aligned} (\text{restoring force}) &= k\, x \\ &= (35\; {\rm N\cdot m^(-1)})\, (100\; {\rm m}) \\ &= 3500\; {\rm N} \end{aligned}`.

Given that
`g = 9.8\; {\rm m\cdot s^(-2)} = 9.8\; {\rm N\cdot kg^(-1)}`, weight of the person would be:

`\begin{aligned}(\text{weight}) &= m\, g \\ &= -(100\; {\rm kg})\, (9.8\; {\rm N\cdot kg^(-1)}) \\ &= -980\; {\rm N}\end{aligned}`.

(Note that since weight points downwards, the sign in front of this force should be negative.)

Assuming that this person is stationary (acceleration is zero,) the resultant force on this person should be zero. In other words:

`(\text{restoring force}) + (\text{weight}) + (\text{unknown force}) = 0`.

Rearrange this equation and solve for the unknown force:

`\begin{aligned} & (\text{unknown force})\\ =\; & - (\text{restoring force}) - (\text{weight}) \\ =\; & - 3500\; {\rm N} - (-980\; {\rm N}) \\ =\; & -2520\; {\rm N} \end{aligned}`.

(Note that the sign in front of this unknown force is negative since this force is supposed to point downwards.)

In other words, a force of magnitude
`2520\; {\rm N}` pointing downwards is required.