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4-2.Friction
hard
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
A$\frac{F}{M}\left( {\cos \,\phi + \mu \,\sin \,\phi } \right) - \mu g$
B$F\,\sin \,\phi /M$
C$\mu F\,\cos \,\phi $
D$\mu F\,\sin \,\phi $
Solution
$R=\mathbf{M g}-F \sin \phi$$\mathrm{Fcos} \phi-\mathrm{f}=\mathrm{Ma}$
or $\mathrm{Fcos} \phi-\mu \mathrm{R}=\mathrm{Ma}$
$\text { Fcos } \phi-\mu(\mathrm{Mg}-\mathrm{F} \sin \phi)=\mathrm{Ma}$
$\therefore \quad a=\frac{F}{M} \cos \phi-\mu\left[g-\frac{F}{M} \sin \phi\right]$
${=\frac{F}{M} \cos \phi+\frac{\mu F}{M} \sin \phi-\mu g}$
${=\frac{F}{M}[\cos \phi+\mu \sin \phi]-\mu g}$
Standard 11
Physics
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