A body of mass 500 g is falling from rest fr | vedclass

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(B) Given: Mass $m = 500 \ g = 0.5 \ kg$, Height $h = 3.2 \ m$, Initial velocity $u = 0 \ ms^{-1}$, Final velocity $v = 6 \ ms^{-1}$, Acceleration due

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$7$

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(B) Given: Mass $m = 500 \ g = 0.5 \ kg$, Height $h = 3.2 \ m$, Initial velocity $u = 0 \ ms^{-1}$, Final velocity $v = 6 \ ms^{-1}$, Acceleration due to gravity $g = 10 \ ms^{-2}$.The initial potential energy of the body at height $h$ is $PE_i = mgh = 0.5 	imes 10 	imes 3.2 = 16 \ J$.The initial kinetic energy is $KE_i = \frac{1}{2}mu^2 = 0 \ J$.Total initial mechanical energy $E_i = PE_i + KE_i = 16 \ J$.The final potential energy at the ground is $PE_f = 0 \ J$.The final kinetic energy is $KE_f = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 0.5 	imes (6)^2 = 0.25 	imes 36 = 9 \ J$.Total final mechanical energy $E_f = PE_f + KE_f = 9 \ J$.The energy lost due to air resistance is the difference between the initial and final mechanical energy:$Loss = E_i - E_f = 16 \ J - 9 \ J = 7 \ J$.

$A$ body of mass $500 \ g$ is falling from rest from a height of $3.2 \ m$ from the ground. If the body reaches the ground with a velocity of $6 \ ms^{-1}$, then the energy lost by the body due to air resistance is (Acceleration due to gravity $= 10 \ ms^{-2}$) (in $J$)

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