Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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?

$$\mathrm{But}\:\mathrm{spring}\:\mathrm{force}\:\mathrm{does}\:\mathrm{not}\:\mathrm{disappear} \\ $$$$\mathrm{immediately},\:\mathrm{then}\:\mathrm{why}\:\mathrm{you}\:\mathrm{have} \\ $$$$\mathrm{neglected}\:{F}_{{x}} \:=\:{kx}\:\mathrm{cos}\:\mathrm{37}°\:\mathrm{due}\:\mathrm{to}\:\mathrm{spring} \\ $$$${B}? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com