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(B) $1$. Resolve the applied force $F$ into horizontal and vertical components: $F_{x} = F \cos 	heta$ and $F_{y} = F \sin 	heta$.$2$. The vertical

Vedclass · Accepted Answer

$\frac{F}{m} \cos 	heta - \mu_{K}(g - \frac{F}{m} \sin 	heta)$

Vedclass · Answer

(B) $1$. Resolve the applied force $F$ into horizontal and vertical components: $F_{x} = F \cos 	heta$ and $F_{y} = F \sin 	heta$.$2$. The vertical forces acting on the block are the normal force $N$ (upwards),the vertical component of the applied force $F \sin 	heta$ (upwards),and the weight $mg$ (downwards). Since there is no vertical motion,the net vertical force is zero: $N + F \sin 	heta = mg$,which gives $N = mg - F \sin 	heta$.$3$. The kinetic friction force $f_{k}$ is given by $f_{k} = \mu_{K} N = \mu_{K}(mg - F \sin 	heta)$.$4$. The net horizontal force acting on the block is the horizontal component of the applied force minus the friction force: $F_{net} = F \cos 	heta - f_{k} = F \cos 	heta - \mu_{K}(mg - F \sin 	heta)$.$5$. Using Newton's second law,$F_{net} = ma$,we get $ma = F \cos 	heta - \mu_{K}(mg - F \sin 	heta)$.$6$. Dividing by mass $m$,the acceleration is $a = \frac{F}{m} \cos 	heta - \mu_{K}(g - \frac{F}{m} \sin 	heta)$.

$A$ block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in the figure. The coefficient of kinetic friction is $\mu_{K}$. Then,the block's acceleration $a$ is given by: ($g$ is acceleration due to gravity)

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