Question

A person is at the top of a tower. He takes a segment of a string which measures 30 cm long when at rest and hooks his 3 kg sword at the end of it. The spring extends to 35 cm long. He will use this spring to get to the ground. The calculated spring constant is 588 N. how much of the spring (measured at equilibrium) does he need in order to have a net force of 0 upon himself when he touches the ground? Assume he hangs the spring from a hook located exactly 30 m above the ground.

Answer 1

29.95 m

Step-by-step explanation:

Let us assume that by "touching the ground", we mean that the person touches the ground with the sword.

In order for the net force on the person to be zero, the spring has to provide the force to lift the sword up instead of the person.

The spring force and the weight must cancel for the net force on the sword to be zero.

How much force does the spring exert on the sword?

`F=k\Delta x`

where Δx = change in spring length.

Now this force must equal the weight of the sword; therefore,

`k\Delta x=mg`

where m = mass of the sword and g = acceleration due to gravity.

Putting in m = 3.0 kg, g = 9.8m/s^2 and k =588 N gives

Solving for Δx gives

`\boxed{\Delta x=0.05m\text{.}}`

Now, if the ground is 30 m away and the spring stretches by 0.05 m, the original (equilibrium) length of the spring must be

`30m-0.05m=29.95m`

Hence, the original length of the spring is 29.95 m.