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Question

A block of mass 5 kg is thrown upwards with a velocity of 10 m/s. It momentarily comes to rest at a height of 4 m. What is the work done by air friction durring this interval?

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Solution

$K i n e t i c e n e r g y = w o r k d o n e b y g r a v i t y + w o r k d o n e b y f r i c t i o n 0.5 (m v^{2}) = m g h + w o r k d o n e b y f r i c t i o n 0.5 * 5 * 10^{2} = 5 * 10 * 4 + w o r k d o n e b y f r i c t i o n 250 - 200 = w o r k d o n e b y f r i c t i o n 50 J o u l e = w o r k d o n e b y f r i c t i o n$

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