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A-1-kg-block-is-being-pushed-against-a-wall-by-a-force-F-75-N-as-shown-in-the-figure-The-coefficient-of-friction-is-0-25-The-magnitude-of-acceleration-of-the-block-is-




Question Number 20579 by Tinkutara last updated on 28/Aug/17
A 1 kg block is being pushed against a  wall by a force F = 75 N as shown in  the figure. The coefficient of friction is  0.25. The magnitude of acceleration of  the block is
$$\mathrm{A}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{is}\:\mathrm{being}\:\mathrm{pushed}\:\mathrm{against}\:\mathrm{a} \\ $$$$\mathrm{wall}\:\mathrm{by}\:\mathrm{a}\:\mathrm{force}\:{F}\:=\:\mathrm{75}\:\mathrm{N}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{figure}.\:\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{is} \\ $$$$\mathrm{0}.\mathrm{25}.\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Commented by ajfour last updated on 28/Aug/17
N=Fcos 37 ° =75×(4/5)=60N  f_(max) =μN=(1/4)×60=15  F_y =Fsin 37°= 75×(3/5)=45N  ΣF_y =ma    F_y −mg−f_(max) =ma  ⇒  45−10−15=a  ⇒   a=20 m/s^2  .
$${N}={F}\mathrm{cos}\:\mathrm{37}\:°\:=\mathrm{75}×\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{60}{N} \\ $$$${f}_{{max}} =\mu{N}=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{60}=\mathrm{15} \\ $$$${F}_{{y}} ={F}\mathrm{sin}\:\mathrm{37}°=\:\mathrm{75}×\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{45}{N} \\ $$$$\Sigma{F}_{{y}} ={ma} \\ $$$$\:\:{F}_{{y}} −{mg}−{f}_{{max}} ={ma} \\ $$$$\Rightarrow\:\:\mathrm{45}−\mathrm{10}−\mathrm{15}={a} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{a}}=\mathrm{20}\:\boldsymbol{{m}}/\boldsymbol{{s}}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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