A 700-N man jumps out of a window into a fire net 10 m below. The net stretches 2 m before bringing the man to rest and tossing him back into the air. What is the spring constant of the net? What is the velocity just before hitting the net? (hint: Use cons. of energy twice)

Weight = m*g

700 = m*g

h = 12m

Ep = m*g*h
Ep = 700 * 12
Ep = 8400 J

Forgot to mention is that the answer for unit is J

To find the spring constant of the net, we need to use the conservation of energy principle. Let's break the problem down into two parts: the man falling towards the net and the net bringing the man to rest and tossing him back.

First, we can use the conservation of energy to find the velocity just before hitting the net:

1. Calculate the gravitational potential energy of the man before falling:
Potential energy = mass x gravity x height
Here, the mass of the man is not given, but the weight (700 N) is given. Weight is equal to mass times gravity, so we can calculate mass using the formula: mass = weight / gravity.
With weight = 700 N and gravity = 9.8 m/s^2, we can calculate the mass of the man.

2. Now, using the potential energy, we can find the velocity just before hitting the net:
Considering only the potential and kinetic energy (ignoring any energy losses due to air resistance or friction), we can write:
Potential energy = Kinetic energy
mgh = (1/2)mv^2
Here, h is the height (10 m), and we have calculated the mass in the previous step.
We can solve this equation for v to find the velocity just before hitting the net.

Next, to find the spring constant, we consider the net bringing the man to rest and tossing him back:

3. After hitting the net, the man is brought to rest, which means all of his kinetic energy is converted into elastic potential energy in the net due to its stretch. We can use the equation for the elastic potential energy stored in a spring:
Elastic potential energy = (1/2)kx^2
Here, x is the stretch or displacement of the net (2 m), and we want to find the spring constant k.
We can rewrite the elastic potential energy equation in terms of the velocity just before hitting the net (v) and the spring constant (k):
(1/2)mv^2 = (1/2)kx^2
Rearranging this equation, we can solve for k.

Using these calculations, we can find the spring constant of the net and the velocity just before hitting the net.

Spring constants cannot have units of J. It has to have force/length units, not force x length.

Neither can Velocities.

The man's mass is

M = 700/g = 71.4 kg

When fully stretched 2 meters, gravitational potential energy of
M*g*(10 + 2) = 117.6 J is converted to potential energy of the net, which is
(1/2) k X^2, where X = 2 meters.
k = 2*117.6/X^2 = 59 N/m

Before hitting the net, hewill have fallen H = 10 meters, so that
V = sqrt(2*g*H) = 14 m/s