Question Number 49097 by peter frank last updated on 02/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
$${T}=\mathrm{3}×\mathrm{10}=\mathrm{30}{N} \\ $$$${Normal}\:{reaction}=\mathrm{5}×\mathrm{10}=\mathrm{50}{N} \\ $$$${limiting}\:{friction}\:{f}_{{L}} =\mathrm{0}.\mathrm{55}×\mathrm{50}=\mathrm{27}.\mathrm{5}{N} \\ $$$${kinetic}\:{friction}\:{f}_{{k}} =\mathrm{0}.\mathrm{30}×\mathrm{50}=\mathrm{15}{N} \\ $$$${appiled}\:{pulling}\:{tension}\:\left(\mathrm{30}{N}\right)>{limiting}\:{friction}\left({f}_{{L}} \:=\mathrm{27}.\mathrm{5}{N}\right) \\ $$$${so}\:{block}\:{will}\:{slide} \\ $$$${accelaration}\:{of}\:{block}=\frac{\mathrm{30}−\mathrm{15}}{\mathrm{5}}=\mathrm{3}{m}/{sec}^{\mathrm{2}} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$