Conservation of Energy EX 6

The system consists of the parachutist and her parachute (total mass of 83 kg). The system is initially at the height of 1000 m and moves at $v$ = 140 km/h = 39 m/s. The parachutist fell through the height of h = 1000 m and therefore the change in her gravitational potential energy is

$\mathrm{\Delta }{U}_G=-(83)(10)(1000)=-8.3\times 10^{5}J$

Her speed decreased from 39 m/s to 7 m/s. This means that she changed her kinetic energy:

$\mathrm{\Delta }K=\frac{1}{2}(83)(7{)}^2-\frac{1}{2}(83)(39{)}^2=-6.1\times 10^{4}J$

There is no spring involved in this problem ($\mathrm{\Delta }{U}_{sp}=0$).

The work done by the air on the system ($W_{air}$) is not zero. We expect that the work done by non-conservative forces is $W_{NC}<0$.

By substituting into the law of conservation of energy $\mathrm{\Delta }K+\mathrm{\Delta }{U}_G+\mathrm{\Delta }{U}_{sp}=W_{NC}$, we get

$(-6.1\times 10^4)+(-8.3\times 10^5)=W_{air}$

The work done by the air on the system is $-8.9\times 10^{5}J$. Therefore the work done by the system on the air is $8.9\times 10^{5}J$.