A 70kg wannabe stuntman at the edge of a roof top jumps vertical up with a speed of 4m/s and falls on a trampoline 2meters below where he started. If the trampoline acts like a Hooke's spring and does a maximum extension of 0.284m going down towards ground, what is the spring constant?

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To find the spring constant, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension. The formula for Hooke's Law is:

F = k * x

Where:
F is the force exerted by the spring,
k is the spring constant,
x is the extension of the spring.

In this case, we have the following information:
Mass of the stuntman (m) = 70 kg
Initial velocity of the stuntman (v) = 4 m/s
Maximum extension of the spring (x) = 0.284 m
Acceleration due to gravity (g) = 9.8 m/s^2

First, let's find the change in gravitational potential energy (ΔPE):
ΔPE = m * g * h

Where:
m is the mass,
g is the acceleration due to gravity,
h is the height (difference in height between the starting point and the trampoline).

ΔPE = 70 kg * 9.8 m/s^2 * 2 m
ΔPE = 1372 J

Next, let's find the change in kinetic energy (ΔKE):
ΔKE = (1/2) * m * v^2

ΔKE = (1/2) * 70 kg * (4 m/s)^2
ΔKE = 560 J

The change in kinetic energy is equal to the change in potential energy, so:

ΔPE = ΔKE
1372 J = 560 J + (1/2) * k * x^2

Rearranging the equation to solve for k:

k = (2 * (ΔPE - ΔKE)) / x^2
k = (2 * (1372 J - 560 J)) / (0.284 m)^2

Calculating the value of k:

k = (2 * 812 J) / (0.284 m)^2
k ≈ 20184 N/m

Therefore, the spring constant is approximately 20184 N/m.