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-requir

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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