An object of mass 24kg is accelerated up a frictionless plane inclined at an angle of 37°. Starting at the bottom from the rest,it covers a distance of 18m in 3secs. a)what is the average power required to accomplish the process? b)what is the instantaneous power required at the end of the 3second interval?


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Question Number 28646 by NECx last updated on 28/Jan/18

$A n o b j e c t o f m a s s 24 k g i s$ $a c c e l e r a t e d u p a f r i c t i o n l e s s p l a n e$ $i n c l i n e d a t a n a n g l e o f 37 ° .$ $S t a r t i n g a t t h e b o t t o m f r o m t h e$ $r e s t, i t c o v e r s a d i s t a n c e o f 18 m$ $i n 3 s e c s .$ $a) w h a t i s t h e a v e r a g e p o w e r$ $r e q u i r e d t o a c c o m p l i s h t h e$ $p r o c e s s ?$ $b) w h a t i s t h e i n s t a n t a n e o u s p o w e r$ $r e q u i r e d a t t h e e n d o f t h e 3 s e c o n d$ $i n t e r v a l ?$
	Answered by ajfour last updated on 28/Jan/18

	$s = \frac{1}{2} {a t}^{2} \Rightarrow a = \frac{2 s}{t^{2}} = \frac{36}{9} = 4 m / s^{2}$ $v = u + a t = 0 + 4 \times 3 = 12 m / s$ $F - m g \sin 37 ° = m a$ $\Rightarrow F = 240 \times \frac{3}{5} + 24 \times 4$ $= 144 + 96 = 240 N$ $P_{a v g} = \frac{W_{F}}{△ t} = \frac{240 \times 18}{3} = 1440 w a t t$ $P_{i n s t} = F v = 240 \times 4 \times 3 = 2880 w a t t .$
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