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A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v . Assume that the magnitude of the acceleration due to gravity is g .

-What is the work (Wd) done on the skydiver, over the distance , by the drag force of the air?

-Find the power (P d) supplied by the drag force after the skydiver has reached terminal velocity v.

Respuesta :

Kazeemsodikisola

Answer:

a. Wd = ½mv²−mgd

b. Pd = -mgv

Explanation:

The skydiver losses potential energy and gains kinetic energy.

Work done by gravity

W=mg

Therefore, work done by gravity is given as

P.E= weight × height

P.E= mg × d

P.E= mgd

The work done by kinetic energy

K.E=½mv²

Then, the drag work becomes

We=K.E-P.E

We=½mv² - mgd

b. Drag power

Power is given as

Power=force × velocity

The force =weight =mg

g is negative since or is against gravity (upward motion)

Pd=F×v

Pd=-mgv

onyebuchinnaji

The work done on the skydiver over the given distance is [tex]W_d = mgd - \frac{1}{2} mv^2[/tex].

The power supplied by the drag force after the skydiver has reached terminal velocity is mgv.

The given parameters;

  • mass of the skydiver, = m
  • distance traveled by the balloon, = d
  • terminal velocity, = v
  • acceleration due to gravity, = g

The work done on the skydiver over the given distance is calculated as follows;

[tex]W_d = \Delta E\\\\W_d = mgd - \frac{1}{2} mv^2[/tex]

The power supplied by the drag force after the skydiver has reached terminal velocity v;

[tex]P_d = Fv\\\\P_d = mgv[/tex]

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