i want of ask in mechanics is this possible?? power=((work done)/(time taken)) if power = p_w , Work= W_d , and time=t ⇒ p_w = (W_d /t) but W_d = force(f)×distance(s) W_d = fs ⇒ p_w = ((f×s)/t) p_w = f × (s/t)(distance/time) p_w = f × velocity but f=ma p_w = m×a×v rearranging ⇒ p_(w ) = m×v×a p_w = momentum(P)×Acceleration(a) Power = momentum × Acceleration. the momentum of a system is directly propotional to the power of that system but will have a minimum momentum when accelerating. what do you think?


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Question Number 59104 by Forkum Michael Choungong last updated on 04/May/19

$i w a n t o f a s k i n m e c h a n i c s i s t h i s$ $p o s s i b l e ? ?$ $p o w e r = \frac{w o r k d o n e}{t i m e t a k e n}$ $i f p o w e r = p_{w}, W o r k = W_{d}, a n d t i m e = t$ $\Rightarrow p_{w} = \frac{W_{d}}{t}$ $b u t W_{d} = f o r c e (f) \times d i s t a n c e (s)$ $W_{d} = f s$ $\Rightarrow p_{w} = \frac{f \times s}{t}$ $p_{w} = f \times \frac{s}{t} (d i s t a n c e / t i m e)$ $p_{w} = f \times v e l o c i t y$ $b u t f = m a$ $p_{w} = m \times a \times v$ $r e a r r a n g i n g$ $\Rightarrow p_{w} = m \times v \times a$ $p_{w} = m o m e n t u m (P) \times A c c e l e r a t i o n (a)$ $P o w e r = m o m e n t u m \times A c c e l e r a t i o n .$ $t h e m o m e n t u m o f a s y s t e m i s d i r e c t l y$ $p r o p o t i o n a l t o t h e p o w e r o f t h a t s y s t e m$ $b u t w i l l h a v e a m i n i m u m m o m e n t u m$ $w h e n a c c e l e r a t i n g .$ $w h a t d o y o u t h i n k ?$
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