A wooden block of mass m resting on a rough s | vedclass

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(A) $1$. Resolve the applied force $F$ into two components: horizontal component $F \cos 	heta$ and vertical component $F \sin 	heta$.$2$. The verti

Vedclass · Accepted Answer

$\frac{F}{m}(\cos 	heta + \mu \sin 	heta) - \mu g$

Vedclass · Answer

(A) $1$. Resolve the applied force $F$ into two components: horizontal component $F \cos 	heta$ and vertical component $F \sin 	heta$.$2$. The vertical forces acting on the block are the normal force $N$ (upward),the vertical component of the applied force $F \sin 	heta$ (upward),and the weight $mg$ (downward). Since there is no vertical motion,$N + F \sin 	heta = mg$,which gives $N = mg - F \sin 	heta$.$3$. The frictional force $f_r$ is given by $f_r = \mu N = \mu(mg - F \sin 	heta)$.$4$. The net horizontal force acting on the block is $F \cos 	heta - f_r$. According to Newton's second law,$F \cos 	heta - f_r = ma$.$5$. Substituting the value of $f_r$: $F \cos 	heta - \mu(mg - F \sin 	heta) = ma$.$6$. Solving for acceleration $a$: $a = \frac{F \cos 	heta - \mu mg + \mu F \sin 	heta}{m} = \frac{F}{m}(\cos 	heta + \mu \sin 	heta) - \mu g$.

$A$ wooden block of mass $m$ resting on a rough surface (coefficient of friction $\mu$) is pulled by a force $F$ at an angle $\theta$ with the horizontal,as shown in the figure. The acceleration of the block moving horizontally is:

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