Question Number 20579 by Tinkutara last updated on 28/Aug/17
$$\mathrm{A}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{is}\:\mathrm{being}\:\mathrm{pushed}\:\mathrm{against}\:\mathrm{a} \\ $$$$\mathrm{wall}\:\mathrm{by}\:\mathrm{a}\:\mathrm{force}\:{F}\:=\:\mathrm{75}\:\mathrm{N}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{figure}.\:\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{is} \\ $$$$\mathrm{0}.\mathrm{25}.\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Commented by ajfour last updated on 28/Aug/17
$${N}={F}\mathrm{cos}\:\mathrm{37}\:°\:=\mathrm{75}×\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{60}{N} \\ $$$${f}_{{max}} =\mu{N}=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{60}=\mathrm{15} \\ $$$${F}_{{y}} ={F}\mathrm{sin}\:\mathrm{37}°=\:\mathrm{75}×\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{45}{N} \\ $$$$\Sigma{F}_{{y}} ={ma} \\ $$$$\:\:{F}_{{y}} −{mg}−{f}_{{max}} ={ma} \\ $$$$\Rightarrow\:\:\mathrm{45}−\mathrm{10}−\mathrm{15}={a} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{a}}=\mathrm{20}\:\boldsymbol{{m}}/\boldsymbol{{s}}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$