A conveyor belt is driven at velocity v by a motor.Sand drops vertically on to the belt at a rate of mkg/s. What is the additional power needed to keep the conveyor belt moving at a steady speed when the sand starts to fall on it? a)(1/2)mv b)mv c)(1/2)mv^2 d)mv^2


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Question Number 49966 by Necxx last updated on 12/Dec/18

$A c o n v e y o r b e l t i s d r i v e n a t v e l o c i t y$ $v b y a m o t o r . S a n d d r o p s v e r t i c a l l y$ $o n t o t h e b e l t a t a r a t e o f m k g / s .$ $W h a t i s t h e a d d i t i o n a l p o w e r$ $n e e d e d t o k e e p t h e c o n v e y o r b e l t$ $m o v i n g a t a s t e a d y s p e e d w h e n$ $t h e s a n d s t a r t s t o f a l l o n i t ?$ $a) \frac{1}{2} m v b) m v c) \frac{1}{2} {m v}^{2} d) {m v}^{2}$
Commented by Necxx last updated on 12/Dec/18

$p l e a s e h e l p s i r s$
	Answered by mr W last updated on 12/Dec/18

	$i n t i m e Δ t s a n d o f m a s s m Δ t d r o p s o n$ $t o t h e b e l t a n d w i l l b e m o v e d i n$ $h o r i z o n t a l d i r e c t i o n f r o m z e r o s p e e d$ $t o s p e e d v . t h e w o r k d o n e t o t h e s a n d i s$ $W = \frac{1}{2} (m Δ t) v^{2}$ $t h e p o w e r n e e d e d i s P = \frac{W}{Δ t} = \frac{1}{2} {m v}^{2}$
	Commented by Necxx last updated on 13/Dec/18

	$t h a n k s b o s s$
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