A parachutist with mass 79.7 kg jumps from an airplane traveling at a speed vi= 112 km/hr at a height H = 2570 m. He lands with a speed of vf = 5.21 m/s. What is the change in mechanical energy (J) of the earth- parachutist system from just after the jump to just before landing?

My work: I started by solving Ug=mgh

(19.7kg)(9.8)(2570) = 20073224.2
Initial Kinetic + 1/2mgv^2
Ug + ki = Ei
final e = 1/2mgv^2f
change in E = Ei-Ef

The problem I am having is with the simple calculations. When I plug in my mass in kg and velocity in m/s.. I am not getting the correct answer in Joules.. Is there some type of conversion that I am missing?

I do not understand your work.

HE started with Kinetic energy and potentail energy. He ended with some kinetic energy.

change= intial energy-final energy
= 1/2 m Vi^2 + mg*h -1/2 m vf^2

where Vi=112000/3600 m/s
Vf= 5.21m/s
m=79.7kg

I have no idea what conversions you might have done.

It seems like you're on the right track with your calculations. However, it's possible that you made a mistake in the conversion of units.

When solving for the potential energy (Ug), you correctly multiplied the mass (79.7 kg), acceleration due to gravity (9.8 m/s^2), and height (2570 m) to get 19,922,054.6 J.

Now, let's calculate the initial kinetic energy (Ki). The initial velocity (vi) given in km/hr needs to be converted to m/s before we can use it in the equation.

vi = 112 km/hr = 112,000 m/3600 s ≈ 31.11 m/s

The initial kinetic energy can be calculated as:

Ki = 0.5 * mass * (vi)^2
= 0.5 * 79.7 kg * (31.11 m/s)^2
≈ 38772.59 J

To calculate the change in mechanical energy (ΔE), we subtract the final kinetic energy (Kf) from the sum of the potential energy (Ug) and the initial kinetic energy (Ki):

Kf = 0.5 * mass * (vf)^2
= 0.5 * 79.7 kg * (5.21 m/s)^2
≈ 1043.59 J

ΔE = (Ug + Ki) - Kf
= (19,922,054.6 J + 38772.59 J) - 1043.59 J
≈ 19,922,783.6 J

So, the change in mechanical energy of the earth-parachutist system from just after the jump to just before landing is approximately 19,922,783.6 Joules.

To determine the change in mechanical energy of the Earth-parachutist system, we need to calculate the initial and final mechanical energies and then find the difference between them.

1. Initial mechanical energy:
The initial mechanical energy (Ei) of the system just after the jump can be calculated by considering the gravitational potential energy (Ug) and the initial kinetic energy (Ki) of the parachutist. The formula for mechanical energy is given by Ei = Ug + Ki.

a) Gravitational potential energy (Ug):
Ug = mgh, where m is the mass (79.7 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (2570 m).
Plugging in the values: Ug = (79.7 kg)(9.8 m/s^2)(2570 m) = 20073224 J.

b) Initial kinetic energy (Ki):
To find the initial kinetic energy, we need to convert the initial velocity (vi) from km/hr to m/s. You correctly mentioned that there might be a conversion issue.

vi = 112 km/hr = (112 km/hr)(1000 m/km)(1 hr/3600 s) ≈ 31.11 m/s

Now we can calculate the initial kinetic energy:
Ki = 1/2 mv^2, where m is the mass (79.7 kg) and v is the velocity (31.11 m/s).
Plugging in the values: Ki = (1/2) (79.7 kg)(31.11 m/s)^2 ≈ 38396.35 J.

Therefore, Ei = Ug + Ki = 20073224 J + 38396.35 J = 20111620.35 J.

2. Final mechanical energy:
We're given the final velocity (vf) of the parachutist, so we can calculate the final kinetic energy (Ef) using the same formula as before.

Ef = 1/2 mvf^2, where m is the mass (79.7 kg) and vf is the velocity (5.21 m/s).
Plugging in the values: Ef = (1/2) (79.7 kg)(5.21 m/s)^2 ≈ 1026.93 J.

3. Change in mechanical energy:
The change in mechanical energy (ΔE) is equal to the difference between the initial and final mechanical energies.

ΔE = Ei - Ef = 20111620.35 J - 1026.93 J ≈ 20110693.42 J.

Therefore, the change in mechanical energy of the Earth-parachutist system from just after the jump to just before landing is approximately 20110693.42 Joules.