A person wants to participate in a bunjee jump. One end of an elastic band 10m long with a spring constant of 50N/m will be attatched to a basket on a crane and the other end to the waist of the person with a mass of 86kg. The person will step off the edge of the basket to be slowed and brought back by the elastic band before hitting the ground (hopefully 2m above) How high should the crane be? (hint: conservation of energy).

Initial PE= Final PE

mgh=mg2+ 1/2 k(h-12)^2
check my thinking.

could i solve it like this?

Fg=Fs
mg=kx
x=mg/k
x=stretch-Eq value
stretch=x+Eq value
than take the stretch value and add 2.

No, the system is accelerating, Fg does not equal Fs except at equilibrium.

so when i plug the givens into your equation it should look like this

(86)(9.8)(?)=(86)(9.8)1/2(50)(h-12)^2

Im confused

so when i plug the givens into your equation it should look like this

(86)(9.8)(?)=(86)(9.8)1/2(50)(h-12)^2

im confused

To find the height the crane should be, we can use the principle of conservation of energy. At the maximum height of the jump, the potential energy is equal to the initial potential energy plus the potential energy gained by stretching the elastic band. We can set up the equation as follows:

Initial potential energy + Potential energy gained by stretching the elastic band = Final potential energy

The initial potential energy of the person is given by the formula: m*g*h, where m is the mass of the person (86 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the crane.

The potential energy gained by stretching the elastic band is given by the formula: (1/2) * k * x^2, where k is the spring constant (50 N/m) and x is the maximum stretch of the elastic band.

The final potential energy is equal to the sum of the person's mass and the potential energy gained by stretching the elastic band at the maximum height (2 meters above the ground).

Now, let's plug in the values into the equation and solve for h:

Initial potential energy + Potential energy gained by stretching the elastic band = Final potential energy

m * g * h + (1/2) * k * x^2 = m * g * (h + 2)

Substituting the given values:

(86 kg) * (9.8 m/s^2) * h + (1/2) * (50 N/m) * (10 m)^2 = (86 kg) * (9.8 m/s^2) * (h + 2)

Simplifying:

843.6 h + 2500 = 843.6 h + 1723.2

Rearranging the equation:

843.6 h - 843.6 h = 1723.2 - 2500

0 h = -776.8

Since the coefficient of h is zero, we cannot solve for a specific height using the given information. The height of the crane required for the person to reach a maximum height of 2 meters above the ground will depend on the initial potential energy stored in the elastic band and the potential energy gained by stretching it.