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Question Number 49093 by peter frank last updated on 02/Dec/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18

	$12) N o r m a l r e a c t i o n N = 5 \times 10 = 50 N$ $l i m i t i n g f r i c t i o n f_{L} = 0.42 \times 50 = 21 N$ $k i n e t i c f r i c t i o n f_{k} = 0.15 \times 50 = 7.5 N$ $a) a p p l i e d f o r c e = 15 N < (f_{L} = 21 N)$ $s o f r i c t i o n a l f o r c e a l s o 15 N$ $n e t f o r c e = 15 N - 15 N = 0$ $h e n c e a c c e l a r a t i o n i s z e r o$ $b) a p p l i e d f o r c e 25 N > (f_{L} = 21 N)$ $n e t f o r c e = 25 N - 7.5 N$ $a c c = \frac{17.5}{5} = 3.5 m e t e r / {s e c}^{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18

	$13) n o r m a l r e a c t i o n = 42 \times 10 = 420 N$ $l i m i t i n g f r i c t i o n f_{L} = 420 \times 0.44 = 184.8$ $l e t a c c e l a t i o n i s a s o p s e u d o f o r c e o n r e f r i g e r a t o r$ $i s 42 a$ $w h e n 42 a ⩽ 184.8 t h e r e f r i g e r a t o r w i l l n o t s l i d e$ $s o a ⩽ \frac{184.8}{42}$ $a ⩽ 4.4 m e t e r / {s e c}^{2} i s p i c k u p a c c e l a r a t i o n$ $p l s c h e c k . . .$
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