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4-2.Friction
hard
In the given figure the acceleration of $M$ is $(g = 10 \,ms^{-2})$
A$\frac{20}{3} \, ms^{-2}$
B$17\, ms^{-2}$
C$\frac{80}{3} \, ms^{-2}$
DNone of these
Solution
$\sin \theta=\frac{3}{5} \quad \Rightarrow \cos \theta=\frac{4}{5}$
$\mathrm{N}=\mathrm{mg}-\mathrm{F} \sin \theta=\frac{10}{3} \times 10-50 \times \frac{3}{5}=\frac{10}{3}$
and $\mathrm{F}+\mathrm{F} \cos \theta-\mu \mathrm{N}=\mathrm{Ma}$
$\Rightarrow \quad 50+50 \times \frac{4}{5}-\frac{1}{3} \times \frac{10}{3}=\frac{10}{3} \mathrm{a}$
$\Rightarrow \quad 90-\frac{10}{9}=\frac{10}{3} \mathrm{a} \Rightarrow \mathrm{a}=\frac{80}{3} \mathrm{m} / \mathrm{s}^{2}$
$\mathrm{N}=\mathrm{mg}-\mathrm{F} \sin \theta=\frac{10}{3} \times 10-50 \times \frac{3}{5}=\frac{10}{3}$
and $\mathrm{F}+\mathrm{F} \cos \theta-\mu \mathrm{N}=\mathrm{Ma}$
$\Rightarrow \quad 50+50 \times \frac{4}{5}-\frac{1}{3} \times \frac{10}{3}=\frac{10}{3} \mathrm{a}$
$\Rightarrow \quad 90-\frac{10}{9}=\frac{10}{3} \mathrm{a} \Rightarrow \mathrm{a}=\frac{80}{3} \mathrm{m} / \mathrm{s}^{2}$
Standard 11
Physics
