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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2415.49N/m
To find the spring constant, we will use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension.
Hooke's Law is given by the equation: F = k * x
Where:
F is the force exerted by the spring
k is the spring constant
x is the extension or displacement of the spring from its equilibrium position
In this case, the maximum extension of the trampoline, x, is given as 0.284m. We need to find the value of the spring constant, k.
To determine the force exerted by the trampoline on the stuntman, we will consider his weight and the upward force exerted by the trampoline.
1. Calculate the weight of the stuntman:
Weight = mass * acceleration due to gravity
Weight = 70kg * 9.8 m/s^2
Weight = 686 N
2. Determine the initial potential energy of the stuntman when he's at the edge of the rooftop:
Potential Energy = mass * gravitational field strength * height
Potential Energy = 70kg * 9.8 m/s^2 * 2m
Potential Energy = 1372 J
3. Determine the maximum potential energy converted to elastic potential energy in the trampoline:
Potential Energy converted = Potential Energy_initial - Potential Energy_final
Potential Energy converted = 1372 J - 0 J (since the stuntman at the lowest point has no potential energy)
Potential Energy converted = 1372 J
4. Determine the force exerted by the trampoline on the stuntman using Hooke's Law:
Force = k * x
We know that potential energy converted = force * distance
So, force = potential energy converted / distance
Substituting the values:
Potential Energy converted = Force * x
1372 J = k * 0.284 m
Solving for k:
k = 1372 J / 0.284 m
k ≈ 4831.69 N/m
Therefore, the spring constant of the trampoline is approximately 4831.69 N/m.