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A 92.0-kg skydiver falls straight downward with an open parachute through a vertical height of 325 m. The skydiver’s velocity remains constant. What is the work done by the nonconservative force of air resistance, which is the only nonconservative force acting?

Respuesta :

fahadisahadam

Answer:

The work done is [tex]293.0kJ[/tex]

Explanation:

Data

  • Mass:   [tex]m=92.0kg[/tex]
  • height:    [tex]h=325m[/tex]
  • acceleration due to gravity:  [tex]g=9.81m/s^{2}[/tex]

Non conservative force act

The work is:

[tex]W=[/tex]Δ[tex]KE[/tex]+Δ[tex]PE[/tex]

KE and PE means kinetic energy and potential energy respectively.

The change in kinetic energy is 0, since the skydiver’s velocity remains constant.

Therefore

[tex]W=0+[/tex]Δ[tex]PE[/tex]

[tex]W=[/tex]Δ[tex]PE[/tex]

[tex]W=mgh[/tex]

substitute the values

[tex]W=92.0kg*9.81m/s^{2} *325m\\W=293020J\\W=293.0kJ[/tex]

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