Airplane and Parachute?

Question

An airplane is flying with a speed of 285 km/h at a height of 3500 m above the ground. A parachutist whose mass is 77.6 kg, jumps out of the airplane, opens the parachute immediately and then lands on the ground with a speed of 3.45 m/s. How much energy was dissipated on the parachute by the air friction? The drag constant of the parachute is 64.0 N/(m/s)^2.

I assume you find the total energy lost but I dont know what to do from there

Okay so my total energy lost is potential energy at top minus kinetic energy on bottum?

Where does the drag constant come into play?

Answer 1

> I assume you find the total energy lost but I dont know what to do from there.

Nothing. That's exactly the answer they're looking for. The energy lost is exactly the energy that was dissipated by the parachute.

To figure that out, Calculate the KE plus PE that the person had when they were sitting in the plane (based on their initial height & initial speed), and compare that to the KE plus PE they had at the moment they touched down (based on their final height and their final speed).

The drag constant is what they call a "red herring" -- a useless fact that is not important in finding the solution. The whole solution can be determined simply from energy considerations.

Answer 2

You do an energy audit: E(H) = P(H) + K(H); where total energy in the parachustist at bail out is E(H) = P(H) + K(H); P(H) = mgH and K(H) = 1/2 mV^2: V = 285 kph, H = 3500 m, m = 77.6 kg.

At landing E(0) = P(0) + K(0) + W; P = 0, K = 1/2 mv^2; where v = 3.45 m/sec W is work to convert P(H) to P(0) and K(H) to K(0) as the parachustist falls...this is the drag force over the H distance to the ground.

Because of the conservation of energy E(H) = P(H) + K(H) = P(0) + K(0) + W = E(0); so that W = P(H) + K(H) - K(0) = mgH + 1/2 m(V^2 - v^2); and W is the energy dissipated by the chute drag (actually the chute/person but the person would be comparatively small).

You have all the numbers for m, g, H, V, and v, you can do the math. Remember to put V in mps rather kph for consistency.

The physics invoked is the conservation of energy. That parachutist has two sources of energy on bail out: potential energy due to the height of the bail out and the kinetic energy of her velocity from the aircraft. That's her total energy, E(H) = P(H) + K(H), at H.

That energy is conserved so it will be the same energy upon hitting the ground; so that E(H) = E(0). But the composition of E(0) will be altered through the work done by the drag force of the chute. Thus, P(H) + K(H) = P(0) + K(0) + W; where W is that drag force acting over the fall of H.

Answer 3

From the speed and altitude the parachutist jumped from, you can derive the potential and kinetic energy.
When reaching the ground, the only kinetic energy left would be the one from descending at 3.45 m/s; ALL the rest of the energy was dissipated by air friction of the parachute (and the parachutist body). In this context, the drag constant of the parachute is useless.