Question

A body of mass 2 kgs falls from a height of 20 m at a plane where acceleration due to gravity is10 m/s^2 velocity 15m/s then work done against air resistance

Answer 1

Answer:

In this type of questions, it's best to apply Work-Energy Theorem : It states that the work done by all the forces will be equal to the change in Kinetic Energy of the energy.

As per Work Energy Theorem :

$work = \Delta(KE)$

$= > W_{gravity} + W_{r} = \dfrac{1}{2} m {v}^{2} - 0$

$= > mgh + W_{r} = \dfrac{1}{2} m {v}^{2} - 0$

$= > (2 \times 10 \times 20) + W_{r} = \dfrac{1}{2} \times 2 \times {(15)}^{2}$

$= > W_{r} = 225 - 400$

$= > W_{r} = - 175 \: J$

So this is the work done by resistive force air. Therefore the work done by operator against air resistance will be of opposite sign (i.e. positive)

So final answer is:

$\boxed{ \red{ \bold{ \huge{W_{r} = + 175 \: J }}}}$

Answer 2

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Question :

A body of mass 2 kgs falls from a height of 20 m at a plane where acceleration due to gravity is10 m/s^2 velocity 15m/s then work done against air resistance.

Solution :

Work Energy Theorem -

$\tt{\red{work = \Delta(KE)work=Δ(KE)}}$

$\tt{\purple{\implies{ W_{gravity} + W_{r} = \dfrac{1}{2} m {v}^{2} }}}$

$\tt{\orange{\implies{mgh + W_{r} = \dfrac{1}{2} m {v}^{2} - 0 }}}$

$</p><p>\tt{\green{\implies{(2 \times 10 \times 20) + W_{r} = \dfrac{1}{2} \times 2 \times {(15)}^{2} }}}$

Solving We Get :

$</p><p>\tt{\boxed{\boxed{ \green{ \bold{ \huge{W_{r} = - 175 \: J }}}}}}$