Question

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How much energy was lost to air friction during this jump?
kJ

process and answer please

Answer 1

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

$E_{pot}-E_{friction}=E_{kin}$

where:

$E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}$

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

$(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]$