A wooden block of mass M resting on a rough h | vedclass

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

Vedclass · Accepted Answer

$\frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g$

Vedclass · Answer

(A) $1$. Resolve the applied force $F$ into horizontal and vertical components: $F \cos \phi$ (horizontal) and $F \sin \phi$ (vertical).$2$. The vertical forces acting on the block are the normal reaction $R$ (upwards),the vertical component of the applied force $F \sin \phi$ (upwards),and the weight $Mg$ (downwards). For vertical equilibrium: $R + F \sin \phi = Mg$,so $R = Mg - F \sin \phi$.$3$. The horizontal force causing motion is $F \cos \phi$. The kinetic friction force $f$ opposing the motion is given by $f = \mu R = \mu(Mg - F \sin \phi)$.$4$. Applying Newton's second law in the horizontal direction: $F \cos \phi - f = Ma$.$5$. Substituting the expression for $f$: $F \cos \phi - \mu(Mg - F \sin \phi) = Ma$.$6$. Solving for acceleration $a$: $Ma = F \cos \phi - \mu Mg + \mu F \sin \phi$.$7$. Dividing by $M$: $a = \frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g$.

$A$ wooden block of mass $M$ resting on a rough horizontal surface is pulled with a force $F$ at an angle $\phi$ with the horizontal. If $\mu$ is the coefficient of kinetic friction between the block and the surface,then the acceleration of the block is:

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