Question Number 225776 by fantastic last updated on 10/Nov/25
Commented by mr W last updated on 10/Nov/25
$${a}_{{x}} \left(\rightarrow\right)={a}_{{y}} \left(\downarrow\right)=\frac{{g}}{\mathrm{2}+{k}+\frac{{M}}{{m}}} \\ $$
Commented by fantastic last updated on 10/Nov/25
$${you}\:{right}. \\ $$$$\:{i}\:{also}\:{got}\:\frac{{mg}}{\mathrm{2}{m}+\mu{m}+{M}} \\ $$
Commented by fantastic last updated on 10/Nov/25
$${a}=\sqrt{{a}_{{x}} ^{\mathrm{2}} +{a}_{{y}} ^{\mathrm{2}} }=\frac{{mg}}{\mathrm{2}{m}+\mu{m}+{M}}\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 10/Nov/25
Commented by fantastic last updated on 11/Nov/25
$${great}\:{sir} \\ $$
Commented by mr W last updated on 10/Nov/25
$${a}_{{y}} ={a}_{{x}} \\ $$$${N}={ma}_{{x}} \\ $$$${ma}_{{y}} ={mg}−{T}−{kN} \\ $$$${ma}_{{x}} ={mg}−{T}−{kma}_{{x}} \\ $$$$\Rightarrow\left(\mathrm{1}+{k}\right){ma}_{{x}} ={mg}−{T}\:\:\:…\left({i}\right) \\ $$$${Ma}_{{x}} ={T}−{N} \\ $$$${Ma}_{{x}} ={T}−{ma}_{{x}} \\ $$$$\Rightarrow\left({M}+{m}\right){a}_{{x}} ={T}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\left({M}+\mathrm{2}{m}+{km}\right){a}_{{x}} ={mg} \\ $$$$\Rightarrow{a}_{{x}} =\frac{{g}}{\mathrm{2}+{k}+\frac{{M}}{{a}}}={a}_{{y}} \\ $$