Question

A body at mass 1 kg falls from rest through
a distance of 200 m and acquires a speed of
50 m/s. Work done against friction of air is
(1) 700
(2) 1250 j
(3) 750 J
(4) 1960 J​

Answer 1

Answer:

work done = change in kinetic enegy

so here work done =

1/2mv^2-0

or 1/2 ×1 50^2 =1250 j

Answer 2

Answer:

(3) 750 J

Explanation:

Given that

mass,m= 1 kg

h= 200 m

v= 50 m/s

From work power energy

Work done due to all forces = Change in kinetic energy.

$W_{gravity}+W_{air\ friction}=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Given that body is in rest initially ,u=0 m/s

$W_{gravity}+W_{air\ friction}=\dfrac{1}{2}mv^2$

Take g=10 m/s²

$1\times 10\times 200+W_{air\ friction}=\dfrac{1}{2}\times 1\times 50^2$

$W_{air\ friction}=\dfrac{1}{2}\times 1\times 50^2-1\times 9.81\times 200$

$W_{air\ friction}=-750\ J$

Negative sign indicates work is against the displacement.

So the option (3) is correct.

(3) 750 J