A bullet of mass 10 ,g is fired horizontally | vedclass

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(B) According to the work-energy theorem, the work done by all forces equals the change in kinetic energy.$W_{gravity} + W_{air} = \Delta K$$mgh + W_{

Vedclass · Accepted Answer

$3755$

Vedclass · Answer

(B) According to the work-energy theorem, the work done by all forces equals the change in kinetic energy.$W_{gravity} + W_{air} = \Delta K$$mgh + W_{air} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$Given: $m = 10 \,g = 0.01 \,kg$, $h = 50 \,m$, $v_i = 1000 \,ms^{-1}$, $v_f = 500 \,ms^{-1}$, $g = 10 \,ms^{-2}$.$W_{gravity} = mgh = 0.01 	imes 10 	imes 50 = 5 \,J$.Change in kinetic energy $\Delta K = \frac{1}{2} 	imes 0.01 	imes (500^2 - 1000^2) = 0.005 	imes (250000 - 1000000) = 0.005 	imes (-750000) = -3750 \,J$.Substituting these values into the work-energy theorem:$5 + W_{air} = -3750$$W_{air} = -3750 - 5 = -3755 \,J$.The work done $AGAINST$ air resistance is the negative of the work done $BY$ air resistance.Work done against air resistance $= -(-3755 \,J) = 3755 \,J$.

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