A body of mass 2 , kg slides down a curved tr | vedclass

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(A) According to the law of conservation of mechanical energy,the potential energy lost by the body is equal to the kinetic energy gained by it.Let $m

Vedclass · Accepted Answer

$4.43$

Vedclass · Answer

(A) According to the law of conservation of mechanical energy,the potential energy lost by the body is equal to the kinetic energy gained by it.Let $m$ be the mass of the body,$g$ be the acceleration due to gravity,$h$ be the height,and $v$ be the final velocity.Here,$m = 2 \, kg$,$h = 1 \, m$,and $g = 9.8 \, m/s^2$.The potential energy at the top is $PE = mgh$.The kinetic energy at the bottom is $KE = \frac{1}{2}mv^2$.Equating the two: $mgh = \frac{1}{2}mv^2$.Canceling $m$ from both sides,we get $gh = \frac{1}{2}v^2$.Therefore,$v = \sqrt{2gh}$.Substituting the values: $v = \sqrt{2 	imes 9.8 	imes 1} = \sqrt{19.6} \approx 4.43 \, m/s$.

$A$ body of mass $2 \, kg$ slides down a curved track which is a quadrant of a circle of radius $1 \, m$. All the surfaces are frictionless. If the body starts from rest,its speed at the bottom of the track is ................. $m/s$.

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