A rain drop (m = 3.35 ｘ 10⁻⁵ kg) falls vertically at a constant speed under the influence of the forces of gravity ad friction. In falling through 100 m, how much work is done by:
(a) Gravity and (b) Friction:
(Ans: (a) 0.0328 J, (b) -0.0328 J )



Given:

Mass of raindrop = m = 3.35 ｘ 10⁻⁵ kg

Height = h = 100 m

Value of g = 9.8 m s⁻²

To Find:

(a) Work done by gravity =  `\W_g` = ?   

(b) Work done by friction =  `\W_f` = ? 


Solution:

(a) Work done by gravity =  `\W_g` = ?   


As the drop fall vertically downwards under the force of gravity so angle = θ = 0°

By using the formula for work done

`\W_g` = `\vec F` . `\vec d`

`\W_g` = F d cos θ  {where F = ma and d = h}

`\W_g` = mgh cos θ

Putting values

`\W_g` = 3.35 ｘ 10⁻⁵ kg ｘ 9.8 m s⁻² ｘ 100 m ｘ cos 0°

`\W_g` = 0.0328 kg m² s⁻²ｘ 1

`\W_g` = 0.0238 N m -----------Ans.(1)



(b) Work done by friction =  `\W_f` = ? 


As the drop fall vertically downwards against the direction of force of friction, so angle = θ = 180°

By using the formula for work done

`\W_f` = `\vec F` . `\vec d`

`\W_f` = F d cos θ  {where F = ma and d = h}

`\W_f` = mgh cos θ

Putting values

`\W_f` = 3.35 ｘ 10⁻⁵ kg ｘ 9.8 m s⁻² ｘ 100 m ｘ cos 180°

`\W_f` = 0.0328 kg m² s⁻²ｘ (-1)

`\W_f` = - 0.0238 N m -----------Ans.(2)

The negative sign shows that work is done against the force of friction.



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