Question

# A block of mass m is kept over another block of mass M and the system rests on a horizontal surface (figure 8-E1). A constant horizontal force F acting on the lower block produces an acceleration $\frac{F}{2\left(m+M\right)}$ in the system, and the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system. Figure

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Solution

## (a) $a=\frac{\mathrm{F}}{2\left(\mathrm{M}+m\right)}\left(\mathrm{given}\right)$ The free-body diagrams of both the blocks are shown below: For the block of mass m, $ma={\mu }_{1}{\mathrm{R}}_{1}\mathrm{and}{\mathrm{R}}_{1}=mg\phantom{\rule{0ex}{0ex}}⇒{\mu }_{1}=\frac{ma}{{\mathrm{R}}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{F}{2\left(\mathrm{M}+m\right)g}$ (b) Frictional force acting on the smaller block, ${f}_{1}={\mu }_{1}{\mathrm{R}}_{1}=\frac{\mathrm{F}}{2\left(\mathrm{M}+m\right)g}×mg\phantom{\rule{0ex}{0ex}}=\frac{m\mathrm{F}}{2\left(\mathrm{M}+m\right)}$ (c) Work done, w = f1s [where s = d] $=\frac{m\mathrm{F}}{2\left(\mathrm{M}+m\right)}×d\phantom{\rule{0ex}{0ex}}=\frac{m\mathrm{F}d}{2\left(\mathrm{M}+m\right)}$

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