A box of mass 4kg is placed on a wooden planck of length 1.5m lying on the ground.The planc is slowly lifted from one end up to a vertical height of 0.5m at which point the box begins to fall down.Calculate coefficient of friction.

when the box just begins to slide $mg\mathrm{cos}\theta =R,mg\mathrm{sin}\theta =\mu Rwhichimpliesthat\phantom{\rule{0ex}{0ex}}mg\mathrm{sin}\theta =\mu mg\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\mu =\mathrm{tan}\theta =\frac{h}{\sqrt{{s}^{2}-{h}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{0.5}{\sqrt{(1.5{)}^{2}-(0.5{)}^{2}}}=\frac{0.5}{\sqrt{2}}=0.3536$

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