A bungee jumper of mass 75kg is stadnding on a platform 53m above a river. The length of the unstretched bungee cord is 11m. The spring costant is 65.6N/m. Calculate the jumper's speed at 19m below the platform on his first fall.
[Also, please state which height you usd and why,and why isn;t Eg used for conservational energy at end.Why is it (Eg= Ek+Ee) and not (Eg=Ek+Ee+Eg), because he still has gravity potential anergy at end doesn't he?]
To calculate the jumper's speed at 19m below the platform on his first fall, we need to consider the conservation of energy.
First, let's determine the initial gravitational potential energy (Eg) when the jumper is standing on the platform. We can calculate this by using the formula Eg = mgh, where m is the mass of the jumper, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the reference point.
Eg = (75 kg)(9.8 m/s^2)(53 m)
Eg = 38535 J
Next, let's determine the elastic potential energy (Ee) stored in the bungee cord at the point where it is unstretched. We can calculate this using the formula Ee = (1/2)kx^2, where k is the spring constant of the bungee cord, and x is the extension or compression length.
Ee = (1/2)(65.6 N/m)(11 m)^2
Ee = 4040.4 J
Since the bungee cord is initially unstretched, we assume that all the gravitational potential energy is converted to elastic potential energy at the lowest point of the jump. Therefore, at the point 19m below the platform, the jumper will have used up all the gravitational potential energy and elastic potential energy.
Now, let's use the conservation of mechanical energy to find the jumper's speed at this point. The conservation of energy equation is given by Eg + Ee = Ek, where Ek represents the kinetic energy.
Since all the potential energy is converted to kinetic energy, we have the equation Eg + Ee = Ek:
38535 J + 4040.4 J = (1/2)mv^2
Simplifying the equation:
42575.4 J = (1/2)(75 kg)v^2
85150.8 J = 75 v^2
v^2 = 1135.34 m^2/s^2
v ≈ 33.7 m/s
Therefore, the jumper's speed at 19m below the platform on his first fall is approximately 33.7 m/s.
Regarding the second part of your question, the equation Eg = Ek + Ee is used because at the lowest point of the jump, all the gravitational potential energy is converted to kinetic energy. There is no remaining gravitational potential energy at that point, so it is not included in the equation.
Additionally, because the jumper is still subject to the force of gravity, even at the lowest point of the jump, it is not necessary to include the gravitational potential energy term (Eg) on the right-hand side of the equation. The gravitational potential energy is already accounted for when calculating the height above the reference point at the beginning of the problem.