First slide

Work Energy Theorem

Question

Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s². The work done by the
(i) gravitational force and the
(ii) resistive force of air is :

Moderate

A

(i) 1 .2s J (ii) -8.2s J
B

(i) 100 J (ii) 8.75 J
C

(i) 10 J (ii) - 8.75 J
D

(i) - 10 J (ii) - 8.25 J

Solution

Work done by the gravity (W_g) = mgh
= 10^-3 x 10x 10³ = 10 J
By work-energy theorem = $= W_{g} + W_{res} = ΔKE$
$10 + W_{res} = \frac{1}{2} \times 10^{- 3} \times (50)^{2}$
W_res = -8.75 J

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