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 $m s^{- 1}$ . Take 'g' constant with a value 10 $m s^{- 2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

Moderate

A

(i) 1.25 J (ii) -8.25 J
B

(i) 100 J (ii) 8.75 J
C

(i) 100 J (ii) -8.75 J
D

(i) -10 J (ii) 8.25 J

Solution

$Here, m = 1 g = 10^{- 3} kg, h = 1 km = 1000 m, v =$ $50 m s^{- 1}, g = 10 m s^{- 2}$
(i) The work done by the gravitational force
$= m g h = 10^{- 3} \times 10 \times 1000 = 10 J$
$(i i) T h e t o t a l w o r k d o n e b y g r a v i t a t i o n a l f o r c e a n d t h e r e s i s t i v e f o r c e o f a i r i s e q u a l t o c h a n g e i n k i n e t i c e n e r g y o f r a i n d r o p .$
$∴ W_{g} + W_{r} = \frac{1}{2} m v^{2} - 0$
$10 + W_{r} = \frac{1}{2} \times 10^{- 3} \times 50 \times 50 or W_{r} = - 8.75 J$

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