Question

A body of mass 1kg falls from rest through the air, a distance of 100 m and acquires a speed of 40m/s . Work done against air friction is

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

m = 1kg.

S = 100m.

v = 40 m/s.

g = 10 m/s².

and,

Let F be the Force applied by air friction.

$\huge{\bold{\underline{Explanation:-}}}$

There are two forces acting on the body.

The two forces are:-

$\large{\bold{ \to Weight \: of \: the \: body \: (mg)}}$

$\large{\bold{ \to Force \: due \: to \: air \: Friction \: (F)}}$

Now,

$\large\bold{Work \: in \: falling \: 10m = Gain \: in\: K.E}$

implies.

$\large{\bold{W = K.E}}$

$\large{ \therefore (mg - F) \times 100 = \frac{1}{2}mv^2}$

Force will be (mg - F) and work done = F.S

Simplifying.

$\large{ \implies mg - F = \frac{1}{200}mv^2}$

$\large{ \implies F = mg - \frac{1}{200}mv^2}$

Substituting the values.

$\large{ \implies F = 1 \times 10 - \frac{1}{ \cancel{200}} \times 1 \times { \cancel{1600}}}$

Now,

$\large{ \implies F = 10 - 8}$

$\large{ \implies F = 2N}$

Now, Work done by air friction.

$\large{\bold{W = F.S \cos \theta}}$

Substituting the values.

$\large{ \implies W = 2 \times 100 \times \cos180 \degree}$

As it will do work against Free Fall of the body.

So, angle will be 180°.

$\large{ \implies W = 200 \times - 1 \: \: \: \: \: \: (\cos180 = - 1)}$

$\huge{\boxed{\boxed{W = - 200 \: J}}}$

So,the work done by the air friction is - 200N.