Question-205685

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Question Number 205685 by MathedUp last updated on 27/Mar/24

Answered by TonyCWX08 last updated on 27/Mar/24

$$\frac{\mathrm{1}}{\mathrm{10}}{g} \\ $$

Commented by MathedUp last updated on 27/Mar/24

$$\mathrm{pls}\:\mathrm{share}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{process} \\ $$

Answered by mr W last updated on 27/Mar/24

Commented by mr W last updated on 27/Mar/24

m_B =mass of object B decrease of potential energy of B is m_B gh increase of kinetic energy of B is ((m_B (v_2 ^2 −v_1 ^2 ))/2) it is given: m_B gh=10×((m_B (v_2 ^2 −v_1 ^2 ))/2) ⇒gh=5(v_2 ^2 −v_1 ^2 ) on the other side v_2 ^2 −v_1 ^2 =2ah ⇒gh=5×2ah ⇒a=(g/(10))=B′s acceleration A and B have the same acceleration, ⇒A′s acceleration is a=(g/(10))

$${m}_{{B}} ={mass}\:{of}\:{object}\:{B} \\ $$$${decrease}\:{of}\:{potential}\:{energy}\:{of}\:{B}\:{is} \\ $$$${m}_{{B}} {gh} \\ $$$${increase}\:{of}\:{kinetic}\:{energy}\:{of}\:{B}\:{is} \\ $$$$\frac{{m}_{{B}} \left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$${it}\:{is}\:{given}: \\ $$$${m}_{{B}} {gh}=\mathrm{10}×\frac{{m}_{{B}} \left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$$\Rightarrow{gh}=\mathrm{5}\left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$${on}\:{the}\:{other}\:{side}\:{v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{ah} \\ $$$$\Rightarrow{gh}=\mathrm{5}×\mathrm{2}{ah}\: \\ $$$$\Rightarrow{a}=\frac{{g}}{\mathrm{10}}={B}'{s}\:{acceleration} \\ $$$${A}\:{and}\:{B}\:{have}\:{the}\:{same}\:{acceleration}, \\ $$$$\Rightarrow{A}'{s}\:{acceleration}\:{is}\:{a}=\frac{{g}}{\mathrm{10}} \\ $$

Commented by MathedUp last updated on 27/Mar/24