Question-24127

Question Number 24127 by math solver last updated on 12/Nov/17

Commented by math solver last updated on 12/Nov/17

$$\mathrm{ques}.\mathrm{2}? \\ $$

Answered by Physics lover last updated on 13/Nov/17

slipping takes place at t=T solving for T : a_(max of 4kg) = ((0.6)/4){20 − 2t Sin 37°} ⇒ a_(max) = (3/(50)) {50 −3t} during slipping a> a_(max) ((2T ∙ (4/5))/6) > (3/(50)){50−3T} ⇒ T > 6.7164 s until that no relative slipping takes place. now at t=3s a_(system ) = ((6 ∙ (4/5))/6) = (4/5)m/s^2 (4kg)((4/5) m/s^2 ) = f_(friction) ⇒ F_(friction ) = ((16)/5) N ⇒ a is correct. note : g was taken to be 10 m/s^2

$${slipping}\:{takes}\:{place}\:{at}\:{t}={T} \\ $$$${solving}\:{for}\:{T}\:: \\ $$$${a}_{{max}\:{of}\:\mathrm{4}{kg}} \:=\:\frac{\mathrm{0}.\mathrm{6}}{\mathrm{4}}\left\{\mathrm{20}\:−\:\mathrm{2}{t}\:{Sin}\:\mathrm{37}°\right\} \\ $$$$\Rightarrow\:{a}_{{max}} \:=\:\frac{\mathrm{3}}{\mathrm{50}}\:\left\{\mathrm{50}\:−\mathrm{3}{t}\right\} \\ $$$${during}\:{slipping} \\ $$$$\:\:{a}>\:{a}_{{max}} \\ $$$$\:\:\frac{\mathrm{2}{T}\:\centerdot\:\frac{\mathrm{4}}{\mathrm{5}}}{\mathrm{6}}\:\:>\:\frac{\mathrm{3}}{\mathrm{50}}\left\{\mathrm{50}−\mathrm{3}{T}\right\} \\ $$$$\Rightarrow\:\:{T}\:>\:\mathrm{6}.\mathrm{7164}\:{s} \\ $$$${until}\:{that}\:{no}\:{relative}\:{slipping} \\ $$$${takes}\:{place}. \\ $$$${now}\:{at}\:{t}=\mathrm{3}{s} \\ $$$${a}_{{system}\:} =\:\:\frac{\mathrm{6}\:\centerdot\:\frac{\mathrm{4}}{\mathrm{5}}}{\mathrm{6}}\:=\:\frac{\mathrm{4}}{\mathrm{5}}{m}/{s}^{\mathrm{2}} \\ $$$$\:\left(\mathrm{4}{kg}\right)\left(\frac{\mathrm{4}}{\mathrm{5}}\:{m}/{s}^{\mathrm{2}} \right)\:=\:{f}_{{friction}} \\ $$$$\Rightarrow\:\:{F}_{{friction}\:} =\:\frac{\mathrm{16}}{\mathrm{5}}\:{N}\:\: \\ $$$$\Rightarrow\:{a}\:{is}\:{correct}. \\ $$$$ \\ $$$${note}\::\:{g}\:{was}\:{taken}\:{to}\:{be}\:\mathrm{10}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$

Commented by math solver last updated on 13/Nov/17