Question

A body of mass m dropped from a height H reaches the ground with a speed of 1.2
$\sqrt{gh}$
the work done by air friction is​

Answer 1

Given

• A body of mass 'm' dropped from a height 'H' and reaches the ground with a speed of 1.2√gh.

To Find

We have to find the work done by air friction.

Solution

A box is dropped from height 'H' means in the air it's initial velocity i.e. u is 0 m/s.

And after reaching the ground it's final velocity i.e. v is 1.2√gh.

Now, by Work-Energy Theorem

Total work done = Total change in Kinetic Energy (∆K.E.)

K.E. = 1/2 mv² - 1/2 mu² (Final change in kinetic energy - Initial change in kinetic energy)

→ W = 1/2 m(1.2√gH)² - 1/2 m(0)²

→ W = 1/2 m(1.44gH)

→ W = 0.72mgH

Also,

Work done by gravity = mgH [(-m)(-gH)]

Now,

Total work done = Work done by gravity + Work done by air friction

Work done by air friction = Total work done - Work done by gravity

→ 0.72mgH - mgH

→ mgH(0.72 - 1)

→ mgH(-0.28)

→ -0.28mgH

Therefore, the work done by air friction is -0.28mgH.

Answer 2

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QUESTION :

A body of mass m dropped from a height H reaches the ground with a speed of 1.2 $\sqrt{gh}$ .

The work done by air friction is ............

PRINCIPLE USED :

....Law of conservation of Mechanical Energy...

According to this Law,

Innitial Mechanical Energy is equal to the Final Mechanical Energy.

SOLUTION : SOLUTION :

According to this question, we have to calculate the work done by Air Friction on this body.

Note that when this body is at height H , it has only potential energy and no kinetic energy and when it reaches the ground, it has only kinetic energy..

Hence ,

LOSs In EnErGy : FiNAL K.E - Innitial P. E

=> [ 1 / 2] m { v } ^2 - m × g × H

=> [ 1 / 2 ] m { 1.2 √ { gh } } ^2 - m × g × h

=> 0.72 m × g × H - m × g × H

=> - 0.28 m g H

=> - 2.8 m H

The work done by Air friction is equal to the loss of energy.

Hence ,

WORk DonE By AiR FriCtiOn = - 2.8 m H......(A)