A particle of mass 2×10^(−27) kg moves according to the following y=5cos(((πt)/3)+(π/4)) find the maximum kinetic energy


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Question Number 40499 by Necxx last updated on 23/Jul/18

$A p a r t i c l e o f m a s s 2 \times 10^{- 27} k g m o v e s$ $a c c o r d i n g t o t h e f o l l o w i n g$ $y = 5 c o s (\frac{π t}{3} + \frac{π}{4})$ $f i n d t h e m a x i m u m k i n e t i c e n e r g y$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

	$\frac{d y}{d t} = 5 . s i n (\frac{Π t}{3} + \frac{Π}{4}) . \frac{Π}{3}$ ${(\frac{d y}{d t})}_{m a x} = \frac{5 Π}{3}$ $k . E = \frac{1}{2} . 2 \times 10^{- 27} . {(\frac{5 Π}{3})}^{2}$
	Commented by Necxx last updated on 23/Jul/18

	$T h a n k y o u v e r y m u c h s i r .$ $I f I d e c i d e d t o c o m p u t e t h e f i n a l$ $a n s w e r d o I n e e d t o c o n v e r t 5 π / 3$ $t o d e g r e e b e f o r e s o l v i n g o r r u n n i n g$ $i t s t r a i g h t j u s t l i k e t h i s$
	Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

	$p u t Π = 3.14$
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