A-body-is-released-from-rest-at-the-top-ofa-plane-inclined-at-30-to-the-horizontaland-4-0m-high-If-the-coefficiet-of-friction-between-the-body-and-plane-is-0-3-calculate-to-the-time-theb-ody-take

Question Number 120365 by help last updated on 30/Oct/20

$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{the}\:\mathrm{top}\: \\ $$$$\mathrm{ofa}\:\mathrm{plane}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{30}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{horizontaland}\:\mathrm{4}.\mathrm{0m}\:\mathrm{high}\:.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{coefficiet}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{body} \\ $$$$\mathrm{an}{d}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{0}.\mathrm{3}.\:\mathrm{calculate}\:\mathrm{to}\:\mathrm{the}\:\mathrm{time}\:\mathrm{theb} \\ $$$$\mathrm{ody}\:\mathrm{takes}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{of}.\:\mathrm{thea} \\ $$

Commented by Dwaipayan Shikari last updated on 30/Oct/20

mgsinθ−f=ma mgsinθ−μmgcosθ=ma g(sinθ−μcosθ)=a t=(√((2d)/a)) =(√((2H)/(sinθ.g(sinθ−μcosθ)))) d=(H/(sinθ)) =(√((32)/(g(1−((3(√3))/(10))))))=(√((320)/(g(10−3(√3)))))

$${mgsin}\theta−{f}={ma} \\ $$$${mgsin}\theta−\mu{mgcos}\theta={ma} \\ $$$${g}\left({sin}\theta−\mu{cos}\theta\right)={a} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{d}}{{a}}}\:=\sqrt{\frac{\mathrm{2}{H}}{{sin}\theta.{g}\left({sin}\theta−\mu{cos}\theta\right)}}\:\:\:\:{d}=\frac{{H}}{{sin}\theta} \\ $$$$=\sqrt{\frac{\mathrm{32}}{{g}\left(\mathrm{1}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{10}}\right)}}=\sqrt{\frac{\mathrm{320}}{{g}\left(\mathrm{10}−\mathrm{3}\sqrt{\mathrm{3}}\right)}} \\ $$$$ \\ $$

Commented by peter frank last updated on 31/Oct/20