A bullet of mass 25,g moving with a velocity | vedclass

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Question

(A) According to the Work-Energy Theorem,the work done by the resistive force is equal to the change in kinetic energy of the bullet.$W = \Delta K$$F

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$10$

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(A) According to the Work-Energy Theorem,the work done by the resistive force is equal to the change in kinetic energy of the bullet.$W = \Delta K$$F \cdot s = \frac{1}{2} m u^2$Given:Mass $m = 25\,g = 25 	imes 10^{-3}\,kg$Velocity $u = 200\,m/s$Distance $s = 5\,cm = 5 	imes 10^{-2}\,m$Substituting the values:$F = \frac{m u^2}{2 s} = \frac{25 	imes 10^{-3} 	imes (200)^2}{2 	imes 5 	imes 10^{-2}}$$F = \frac{25 	imes 10^{-3} 	imes 40000}{10 	imes 10^{-2}}$$F = \frac{1000}{0.1} = 10000\,N$$F = 10\,kN$.

$A$ bullet of mass $25\,g$ moving with a velocity of $200\,m/s$ is stopped within $5\,cm$ of the target. The average resistance offered by the target is ............... $kN$.

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