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(D) Given: Mass $m = 2 \ kg$,acceleration $a = 4 \ m/s^2$,angle $	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Case $1$: Body sliding down.The component of gr

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$20$

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(D) Given: Mass $m = 2 \ kg$,acceleration $a = 4 \ m/s^2$,angle $	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Case $1$: Body sliding down.The component of gravitational force along the plane is $mg \sin 	heta = 2 	imes 10 	imes \sin 30^{\circ} = 20 	imes 0.5 = 10 \ N$.Let $f$ be the frictional force acting upwards. The equation of motion is:$mg \sin 	heta - f = ma$$10 - f = 2 	imes 4$$10 - f = 8$$f = 2 \ N$.Case $2$: Body moving up with the same acceleration.Let $F$ be the external force applied upwards. The frictional force $f$ will now act downwards.The equation of motion is:$F - mg \sin 	heta - f = ma$$F - 10 - 2 = 2 	imes 4$$F - 12 = 8$$F = 20 \ N$.

$A$ body of mass $2 \ kg$ slides down with an acceleration of $4 \ m/s^2$ on an inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be (Acceleration due to gravity $= 10 \ m/s^2$) (in $N$)

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