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Question Number 20989 by Tinkutara last updated on 09/Sep/17

$$\mathrm{In}\mathrm{the}\mathrm{figure}\mathrm{shown}\mathrm{below},\mathrm{the}\mathrm{block}\mathrm{of}$$$$\mathrm{mass}2\mathrm{kg}\mathrm{is}\mathrm{at}\mathrm{rest}.\mathrm{If}\mathrm{the}\mathrm{spring}\mathrm{constant}$$$$\mathrm{of}\mathrm{both}\mathrm{the}\mathrm{springs}A\mathrm{and}B\mathrm{is}100\mathrm{N}/\mathrm{m}$$$$\mathrm{and}\mathrm{spring}B\mathrm{is}\mathrm{cut}\mathrm{at}t=0,\mathrm{then}$$$$\mathrm{magnitude}\mathrm{of}\mathrm{acceleration}\mathrm{of}\mathrm{block}$$$$\mathrm{immediately}\mathrm{is}$$

Commented by Tinkutara last updated on 09/Sep/17

Answered by dioph last updated on 09/Sep/17

$$t=0^{-}$$$$2kx\sin 37°=mg$$$$x=\frac{mg}{2k\sin 37°}\cong \frac{2\times 9.8}{2\times 100\times 0.6}\cong 0.163\mathrm{m}$$$$t=0^{+}$$$$F_x=kx\cos 37°\cong 100\times 0.163\times 0.8$$$$F_x\cong 13.067\mathrm{N}$$$$F_y=mg-kx\sin 37°$$$$F_y\cong 9.8\mathrm{N}$$$$a=\frac{\sqrt{F_x^2+F_y^2}}{m}\cong \frac{16.334}{2}$$$$a\cong 8.167\mathrm{m}/{\mathrm{s}}^2$$

Commented by Tinkutara last updated on 13/Sep/17

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

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