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

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Question

$R=\mathbf{M g}-F \sin \phi$

$\mathrm{Fcos} \phi-\mathrm{f}=\mathrm{Ma}$

or $\mathrm{Fcos} \phi-\mu \mathrm{R}=\mathrm{Ma}$

$	ext { Fcos } \phi-\m

Vedclass · Accepted Answer

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

Vedclass · Answer

$R=\mathbf{M g}-F \sin \phi$

$\mathrm{Fcos} \phi-\mathrm{f}=\mathrm{Ma}$

or $\mathrm{Fcos} \phi-\mu \mathrm{R}=\mathrm{Ma}$

$	ext { Fcos } \phi-\mu(\mathrm{Mg}-\mathrm{F} \sin \phi)=\mathrm{Ma}$

$	herefore \quad a=\frac{F}{M} \cos \phi-\mu\left[g-\frac{F}{M} \sin \phiight]$

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

${=\frac{F}{M}[\cos \phi+\mu \sin \phi]-\mu g}$

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 acceleration of the block is