Question-74415

Question Number 74415 by Learner-123 last updated on 24/Nov/19

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

$${please}\:{solve}\:{this}\:{by}\:{conservation}\:{of} \\ $$$${energy}. \\ $$

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 .

$$\frac{\mathrm{1}}{\mathrm{2}}{k}\left\{\left({s}_{\mathrm{0}} +\mathrm{2}{x}\right)^{\mathrm{2}} −{s}_{\mathrm{0}} ^{\mathrm{2}} \right\}+\frac{\mathrm{1}}{\mathrm{2}}\left({m}+{M}\right){v}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({m}\rho^{\mathrm{2}} \right)\left(\frac{{v}}{{r}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\left({M}+{m}\right){gx}\:\:\:\:\:\:\:\:\:\:\:\Rightarrow \\ $$$$\:{v}^{\mathrm{2}} \left\{\left({m}+{M}\right)+{m}\left(\frac{\rho}{{r}}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({M}+{m}\right){gx}−\mathrm{4}{kx}\left({x}+{s}_{\mathrm{0}} \right) \\ $$$${Taking}\:\:{g}=\mathrm{10}\left({m}/{s}^{\mathrm{2}} \right) \\ $$$${v}^{\mathrm{2}} \left(\mathrm{150}+\mathrm{50}×\frac{\mathrm{9}}{\mathrm{16}}\right)=\mathrm{3000}×\frac{\mathrm{1}}{\mathrm{20}}−\mathrm{300}×\frac{\mathrm{150}}{\mathrm{1000}} \\ $$$$\Rightarrow\:\:{v}^{\mathrm{2}} =\:\left(\frac{\mathrm{105}×\mathrm{16}}{\mathrm{2850}}\right)\:\frac{{m}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{v}=\sqrt{\frac{\mathrm{21}×\mathrm{16}}{\mathrm{570}}}\:\frac{{m}}{{s}}\:\approx\:\mathrm{0}.\mathrm{77}\:{m}/{s}\:. \\ $$

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

$${thanks}\:{sir}! \\ $$

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