Question

An object of mass 4 kg falls from rest through a vertical distance of 20 m and reaches with velocity of 10 (m)//(s) on ground. How much work is done by air friction?

Answer 1

Given:

• Mass of the object = 4kg

•Vertical distance (h)=20m

• Velocity = 10m/s

To find :

• Work done by air friction =?

$\huge{\mathfrak{ \red{solution}}}$

As we know that , By work-energy Theorem:

• Work done by all forces = Change in kinetic energy

$\rightarrow \: W_mg + W_a = \frac{1}{2} m {v}^{2} \\ \ \\ \rightarrow \: W_a = \frac{1}{2} m {v}^{2} - W_mg \\ \\ \rightarrow \: W_a = \frac{1}{2} m {v}^{2} - mgh \\ \\ \rightarrow \: W_a = \frac{1}{2} (4)( {10}^{2} ) - 4(10)(20) \\ \\ \rightarrow \: W_a = 200 - 800 \\ \\ \rightarrow \: W_a = - 600 \: J \:$

Hence ,the work done by air friction is

$\huge\boxed{ \mathfrak{ \bold{ \green{ - 600J}}}}$

Answer 2

$\huge{\underline{\underline{\red{Answer}}}}$

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$\underline{\mathbb{\green{GIVEN}}}$

• mass= 4kg
• u=0m/s
• h=20m
• v or g=10m/s

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$\underline{\mathbb{\green{NEED\:TO\:FIND}}}$

》 The Negative Work Done by Air=?

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$\bf{\underline{\underline{\red{SOLUTION}}}}$

In absence of any air friction, the velocity should be-

$= \sqrt{2gh} \\ = \sqrt{2 \times 10 \times 20}m/s \\ = \sqrt{20 \times 20} m/s \\ = 20m/s$

This should be the ideal velocity in absence of any friction.

Thus

$\underline{\boxed{\pink{ideal\:velocity=20m/s}}}$

But given Velocity=10m/s

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Thus the work done by friction is

$\frac{1}{2} m {v_{ideal}}^{2} - \frac{1}{2}m {v_{actual}}^{2} \\ = \frac{1}{2} \times 4 \times ( {20}^{2} - {10}^{2} )Joules \\ = 2 \times (20 + 10) \times (20 - 10)Joules \\ = 2 \times 30 \times 10Joules \\ = 600Joules$

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$\bold{\underline{\pink{as\:The\:work\:done\:is}}}$

$\large{\pink{opppsite\:to\:direction\:of\:motion,}}$

$\large{\boxed{\underline{\red{Work\:done=(-600)Joules}}}}$