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Question Number 24127 by math solver last updated on 12/Nov/17

Commented by math solver last updated on 12/Nov/17

$ques . 2 ?$
	Answered by Physics lover last updated on 13/Nov/17
	$slipping takes place at t=T solving for T : a_(max of 4kg) = ((0.6)/4){20 − 2t Sin 37°} ⇒ a_(max) = (3/(50)) {50 −3t} during slipping a> a_(max) ((2T ∙ (4/5))/6) > (3/(50)){50−3T} ⇒ T > 6.7164 s until that no relative slipping takes place. now at t=3s a_(system ) = ((6 ∙ (4/5))/6) = (4/5)m/s^2 (4kg)((4/5) m/s^2 ) = f_(friction) ⇒ F_(friction ) = ((16)/5) N ⇒ a is correct. note : g was taken to be 10 m/s^2$
	$s l i p p i n g t a k e s p l a c e a t t = T$ $s o l v i n g f o r T :$ $a_{m a x o f 4 k g} = \frac{0.6}{4} {20 - 2 t S i n 37 °}$ $\Rightarrow a_{m a x} = \frac{3}{50} {50 - 3 t}$ $d u r i n g s l i p p i n g$ $a > a_{m a x}$ $\frac{2 T \cdot \frac{4}{5}}{6} > \frac{3}{50} {50 - 3 T}$ $\Rightarrow T > 6.7164 s$ $u n t i l t h a t n o r e l a t i v e s l i p p i n g$ $t a k e s p l a c e .$ $n o w a t t = 3 s$ $a_{s y s t e m} = \frac{6 \cdot \frac{4}{5}}{6} = \frac{4}{5} m / s^{2}$ $(4 k g) (\frac{4}{5} m / s^{2}) = f_{f r i c t i o n}$ $\Rightarrow F_{f r i c t i o n} = \frac{16}{5} N$ $\Rightarrow a i s c o r r e c t .$ $n o t e : g w a s t a k e n t o b e 10 m / s^{2}$
	Commented by math solver last updated on 13/Nov/17

	$correct bro! :)$
	Commented by Physics lover last updated on 13/Nov/17

	$i f y o u d o i t t h i s w a y, y o u w i l l n o t$ $h a v e t o m e m o r i z e t h e f o r m u l a .$
	Commented by Physics lover last updated on 13/Nov/17

	$c a n y o u t r y Q 24144, i t s a n e a s y$ $o n e .$
	Commented by math solver last updated on 13/Nov/17

	$ok i will try . .$
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