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Question Number 78161 by TawaTawa last updated on 14/Jan/20

Answered by mr W last updated on 15/Jan/20

M=5 kg m=2 kg T−μmg=ma F−T−μmg−μ(m+Mg)=Ma F−μ(3m+M)g=(m+M)a ⇒a=((F−μ(3m+M)g)/(m+M))=((80−0.44×11×10)/7) =4.51 m/s^2

$${M}=\mathrm{5}\:{kg} \\ $$$${m}=\mathrm{2}\:{kg} \\ $$$${T}−\mu{mg}={ma} \\ $$$${F}−{T}−\mu{mg}−\mu\left({m}+{Mg}\right)={Ma} \\ $$$${F}−\mu\left(\mathrm{3}{m}+{M}\right){g}=\left({m}+{M}\right){a} \\ $$$$\Rightarrow{a}=\frac{{F}−\mu\left(\mathrm{3}{m}+{M}\right){g}}{{m}+{M}}=\frac{\mathrm{80}−\mathrm{0}.\mathrm{44}×\mathrm{11}×\mathrm{10}}{\mathrm{7}} \\ $$$$=\mathrm{4}.\mathrm{51}\:{m}/{s}^{\mathrm{2}} \\ $$

Commented by TawaTawa last updated on 15/Jan/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by TawaTawa last updated on 15/Jan/20

I appreciate

$$\mathrm{I}\:\mathrm{appreciate} \\ $$

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