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In-the-figure-shown-below-the-block-of-mass-2-kg-is-at-rest-If-the-spring-constant-of-both-the-springs-A-and-B-is-100-N-m-and-spring-B-is-cut-at-t-0-then-magnitude-of-acceleration-of-block-immedi

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Question Number 20989 by Tinkutara last updated on 09/Sep/17
In the figure shown below, the block of mass 2 kg is at rest. If the spring constant of both the springs A and B is 100 N/m and spring B is cut at t = 0, then magnitude of acceleration of block immediately is
$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown}\:\mathrm{below},\:\mathrm{the}\:\mathrm{block}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{constant} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{the}\:\mathrm{springs}\:{A}\:\mathrm{and}\:{B}\:\mathrm{is}\:\mathrm{100}\:\mathrm{N}/\mathrm{m} \\ $$$$\mathrm{and}\:\mathrm{spring}\:{B}\:\mathrm{is}\:\mathrm{cut}\:\mathrm{at}\:{t}\:=\:\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{magnitude}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{block} \\ $$$$\mathrm{immediately}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 09/Sep/17
Answered by dioph last updated on 09/Sep/17
t = 0^− 2kx sin 37° = mg x = ((mg)/(2k sin 37°)) ≅ ((2×9.8)/(2×100 × 0.6)) ≅ 0.163 m t = 0^+ F_x = kx cos 37° ≅ 100×0.163×0.8 F_x ≅ 13.067 N F_y = mg − kx sin 37° F_y ≅ 9.8 N a = ((√(F_x ^2 +F_y ^2 ))/m) ≅ ((16.334)/2) a ≅ 8.167 m/s^2
$${t}\:=\:\mathrm{0}^{−} \\ $$$$\mathrm{2}{kx}\:\mathrm{sin}\:\mathrm{37}°\:=\:{mg} \\ $$$${x}\:=\:\frac{{mg}}{\mathrm{2}{k}\:\mathrm{sin}\:\mathrm{37}°}\:\cong\:\frac{\mathrm{2}×\mathrm{9}.\mathrm{8}}{\mathrm{2}×\mathrm{100}\:×\:\mathrm{0}.\mathrm{6}}\:\cong\:\mathrm{0}.\mathrm{163}\:\mathrm{m} \\ $$$${t}\:=\:\mathrm{0}^{+} \\ $$$${F}_{{x}} \:=\:{kx}\:\mathrm{cos}\:\mathrm{37}°\:\cong\:\mathrm{100}×\mathrm{0}.\mathrm{163}×\mathrm{0}.\mathrm{8} \\ $$$${F}_{{x}} \:\cong\:\mathrm{13}.\mathrm{067}\:\mathrm{N} \\ $$$${F}_{{y}} \:=\:{mg}\:−\:{kx}\:\mathrm{sin}\:\mathrm{37}° \\ $$$${F}_{{y}} \:\cong\:\mathrm{9}.\mathrm{8}\:\mathrm{N} \\ $$$${a}\:=\:\frac{\sqrt{{F}_{{x}} ^{\mathrm{2}} +{F}_{{y}} ^{\mathrm{2}} }}{{m}}\:\cong\:\frac{\mathrm{16}.\mathrm{334}}{\mathrm{2}} \\ $$$${a}\:\cong\:\mathrm{8}.\mathrm{167}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 13/Sep/17
But spring force does not disappear immediately, then why you have neglected F_x = kx cos 37° due to spring B?
$$\mathrm{But}\:\mathrm{spring}\:\mathrm{force}\:\mathrm{does}\:\mathrm{not}\:\mathrm{disappear} \\ $$$$\mathrm{immediately},\:\mathrm{then}\:\mathrm{why}\:\mathrm{you}\:\mathrm{have} \\ $$$$\mathrm{neglected}\:{F}_{{x}} \:=\:{kx}\:\mathrm{cos}\:\mathrm{37}°\:\mathrm{due}\:\mathrm{to}\:\mathrm{spring} \\ $$$${B}? \\ $$

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