Question

A bullet of 50 gram strikes a target with a velocity of 600 M per second its velocity is reduced to 150 M per second after penetrating the target how much energy of bullet was used to penetrate the target

Answer 1

energy = (1/2)×0.050kg×(600^2-150^2)
= 8437.5 J.

thanks for asking.

Answer 2

Given :

Mass of bullet , m = 50 gm = 0.05 m .

Initial velocity of bullet , u = 600 m/s .

Velocity of bullet after penetrating , v = 150 m/s .

To Find :

Energy of bullet used to penetrate the target .

Solution :

We know , by conservation of energy .

Energy used = Loss in kinetic energy

$Energy \ used = K.E_i-K.E_f\\\\Energy \ used = \dfrac{1}{2}mu^2-\dfrac{1}{2}mv^2\\\\Energy \ used =\dfrac{m}{2}(u^2-v^2)\\\\Energy \ used =\dfrac{0.05}{2}\times (600^2-150^2)\\\\Energy \ used =8437.5\ J$

Learn More :

Conservation of energy

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