Menu Close

A-wedge-has-a-weight-of-9kg-And-a-block-has-a-weight-of-2kg-If-the-block-starts-sliding-with-an-angle-of-45-with-the-horizontal-then-what-is-accelaration-of-the-wedge-




Question Number 102753 by Dwaipayan Shikari last updated on 11/Jul/20
A wedge has a weight of 9kg . And a block has a weight of 2kg  If the block starts sliding with an angle of 45° with the  horizontal then what is accelaration of the wedge?
$${A}\:{wedge}\:{has}\:{a}\:{weight}\:{of}\:\mathrm{9}{kg}\:.\:{And}\:{a}\:{block}\:{has}\:{a}\:{weight}\:{of}\:\mathrm{2}{kg} \\ $$$${If}\:{the}\:{block}\:{starts}\:{sliding}\:{with}\:{an}\:{angle}\:{of}\:\mathrm{45}°\:{with}\:{the} \\ $$$${horizontal}\:{then}\:{what}\:{is}\:{accelaration}\:{of}\:{the}\:{wedge}? \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 11/Jul/20
N=mg cos θ−ma sin θ  Ma=N sin θ  Ma=(mg cos θ−ma sin θ) sin θ  (M+m sin^2  θ)a=mg sin θcos θ  ⇒a=((mg sin θ cos θ)/(M+m sin^2  θ))  =((2 g×(1/2))/(9+2×(1/2)))=(g/(10))
$${N}={mg}\:\mathrm{cos}\:\theta−{ma}\:\mathrm{sin}\:\theta \\ $$$${Ma}={N}\:\mathrm{sin}\:\theta \\ $$$${Ma}=\left({mg}\:\mathrm{cos}\:\theta−{ma}\:\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\theta \\ $$$$\left({M}+{m}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right){a}={mg}\:\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}=\frac{{mg}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{{M}+{m}\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{\mathrm{2}\:{g}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{9}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}=\frac{{g}}{\mathrm{10}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *