imported>Patrick: Created page with '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. T…'

2011-08-18T17:15:50Z

Created page with '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 <math>v</math> = 140 km/h = 39 m/s. T…'

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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 <math>v</math> = 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

<math>\Delta U_G = -(83)(10)(1000) = -8.3 \times 10^5 J</math>
 
Her speed decreased from 39 m/s to 7 m/s. This means that she changed her kinetic energy:

<math>\Delta K = \frac{1}{2}(83)(7)^2 - \frac{1}{2}(83)(39)^2 = -6.1 \times 10^4 J</math>
 
There is no spring involved in this problem (<math>\Delta U_{sp} = 0</math>).

The work done by the air on the system (<math>W_{air}</math>) is not zero. We expect that the work done by non-conservative forces is <math>W_{NC} < 0</math>.
 
By substituting into the law of conservation of energy <math>\Delta K + \Delta U_G + \Delta U_{sp} = W_{NC}</math>, we get

<math>(-6.1 \times 10^4) + (-8.3 \times 10^5) = W_{air}</math>
 
The work done by the air on the system is <math>-8.9 \times 10^5 J</math>. Therefore '''the work done by the system on the air is <math>8.9 \times 10^5 J</math>'''.

Conservation of Energy EX 6 - Revision history

imported>Patrick: Created page with '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. T…'

imported>Patrick: Created page with '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. T…'