An airplane is flying with a speed of 250km/h at a height of 3610m above the ground. A parachutist whose mass is 75.2kg, jumps out of the airplane, opens the parachute and then lands on the ground with a speed of 3.75m/s . How much energy was dissipated on the parachute by the air friction?
So I assumed the horizontal speed of 250km/h of the airplane is not needed to obtain the solution?
Using Conservation of Energy theorem:
Epgi+Eki = Epgf+Ekf+E
mghi+(1/2)mvi^2 = mghf+(1/2)mvf^2+E
(75.2kg)(9.80m/s^2)(3610m)+(1/2)(75.2kg)(0m/s)^2 = (75.2kg)(9.80m/s^2)(0m)+(1/2)(75.2kg)(3.75m/s)^2+E
2660000J+0J = 0J+529J+E
E = 2660000J-529J
= 2660000J
(The full answer I got without rounding was 2659896.85J), but apparently this is incorrect? How?
His initial KE was dependent on air speed,
Initial energy=energy dissipated+final KE
1/2 mvi^2+mgh=energy dissipated +final KE.
In yours, you assumed vi=0, not true. It is the initial airspeed. You have to assume the parachute absorved it.
Thank you. But I am still confused about why exactly the airplane's speed affects the parachutist's initial kinetic energy. The way I break it down is that the airplane's speed is horizontal and the parachutist's initial kinetic energy deals with vertical components.
KE is NOT A Vector, nor is it directional.
To calculate the energy dissipated on the parachute by air friction, we can use the principle of conservation of energy.
The initial energy of the parachutist is given by the sum of their kinetic energy (KE) and potential energy (PE) while in the airplane. The final energy is the sum of their kinetic energy when they land on the ground. The difference between the initial and final energies gives us the energy dissipated on the parachute.
First, let's calculate the initial energy (E_initial) of the parachutist:
E_initial = KE_initial + PE_initial
The parachutist's kinetic energy (KE_initial) is given by:
KE_initial = 1/2 * m * v^2
Where:
m = mass of the parachutist = 75.2 kg
v = velocity of the parachutist while in the airplane = 250 km/h = (250 * 1000) / 3600 m/s
Next, the potential energy (PE_initial) is given by:
PE_initial = m * g * h
Where:
g = acceleration due to gravity = 9.8 m/s^2
h = height of the airplane above the ground = 3610 m
Now, let's calculate the final kinetic energy (KE_final) of the parachutist when they land on the ground:
KE_final = 1/2 * m * v_final^2
Where:
v_final = velocity of the parachutist when they land on the ground = 3.75 m/s
The energy dissipated on the parachute (E_dissipated) is given by:
E_dissipated = E_initial - E_final = KE_initial + PE_initial - KE_final
Now we can substitute the given values into the equations to calculate the energy dissipated on the parachute.