A body of weight 64 ~N is pushed with just en | vedclass

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(D) Given: Weight $W = 64 ~N$, coefficient of static friction $\mu_s = 0.6$, coefficient of kinetic friction $\mu_k = 0.4$.To start the motion, the ap

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$0.2 ~g$

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(D) Given: Weight $W = 64 ~N$, coefficient of static friction $\mu_s = 0.6$, coefficient of kinetic friction $\mu_k = 0.4$.To start the motion, the applied force $F$ must be equal to the limiting friction: $F = \mu_s N = \mu_s W$.Substituting the values: $F = 0.6 	imes 64 ~N = 38.4 ~N$.Once the body is in motion, the kinetic friction acting on it is $f_k = \mu_k N = \mu_k W$.Substituting the values: $f_k = 0.4 	imes 64 ~N = 25.6 ~N$.The net force acting on the body is $F_{net} = F - f_k = (\mu_s - \mu_k) W$.Using Newton's second law, $F_{net} = ma$, where $m = \frac{W}{g}$.So, $(\mu_s - \mu_k) W = \frac{W}{g} 	imes a$.$a = (\mu_s - \mu_k) g$.$a = (0.6 - 0.4) g = 0.2 ~g$.

List-$I$	List-$II$
$A$. Butane $\to$ Isobutane	$a$. Cracking
$B$. Butane $\to$ Lower hydrocarbons	$b$. Isomerisation
$C$. $n$-Heptane $\to$ Toluene	$c$. Reed reaction
$D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$	$d$. Aromatisation

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