Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
Question Number 74415 by Learner-123 last updated on 24/Nov/19

Commented by Learner-123 last updated on 24/Nov/19

$p l e a s e s o l v e t h i s b y c o n s e r v a t i o n o f$ $e n e r g y .$
	Answered by ajfour last updated on 24/Nov/19
	$(1/2)k{(s_0 +2x)^2 −s_0 ^2 }+(1/2)(m+M)v^2 +(1/2)(mρ^2 )((v/r))^2 = (M+m)gx ⇒ v^2 {(m+M)+m((ρ/r))^2 } = 2(M+m)gx−4kx(x+s_0 ) Taking g=10(m/s^2 ) v^2 (150+50×(9/(16)))=3000×(1/(20))−300×((150)/(1000)) ⇒ v^2 = (((105×16)/(2850))) (m^2 /s^2 ) v=(√((21×16)/(570))) (m/s) ≈ 0.77 m/s .$
	$\frac{1}{2} k {{(s_{0} + 2 x)}^{2} - s_{0}^{2}} + \frac{1}{2} (m + M) v^{2} + \frac{1}{2} (m ρ^{2}) {(\frac{v}{r})}^{2}$ $= (M + m) g x \Rightarrow$ $v^{2} {(m + M) + m {(\frac{ρ}{r})}^{2}}$ $= 2 (M + m) g x - 4 k x (x + s_{0})$ $T a k i n g g = 10 (m / s^{2})$ $v^{2} (150 + 50 \times \frac{9}{16}) = 3000 \times \frac{1}{20} - 300 \times \frac{150}{1000}$ $\Rightarrow v^{2} = (\frac{105 \times 16}{2850}) \frac{m^{2}}{s^{2}}$ $v = \sqrt{\frac{21 \times 16}{570}} \frac{m}{s} \approx 0.77 m / s .$
	Commented by Learner-123 last updated on 24/Nov/19

	$t h a n k s s i r!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum