The average force necessary to stop a bullet | vedclass

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Question

(D) Given: Mass of the bullet $m = 20\, g = 0.02\, kg$.Initial velocity $u = 250\, m/s$.Final velocity $v = 0\, m/s$ (as it stops).Distance penetrated

Vedclass · Accepted Answer

$5.2 	imes 10^3\, N$

Vedclass · Answer

(D) Given: Mass of the bullet $m = 20\, g = 0.02\, kg$.Initial velocity $u = 250\, m/s$.Final velocity $v = 0\, m/s$ (as it stops).Distance penetrated $s = 12\, cm = 0.12\, m$.Using the work-energy theorem or the third equation of motion $v^2 = u^2 + 2as$:$0^2 = (250)^2 + 2 	imes a 	imes 0.12$$a = -\frac{62500}{0.24} = -260416.67\, m/s^2$.The magnitude of the average force is $F = m|a|$.$F = 0.02 	imes 260416.67 = 5208.33\, N$.Rounding to the nearest given option,$F \approx 5.2 	imes 10^3\, N$.

The average force necessary to stop a bullet of mass $20\, g$ moving with a speed of $250\, m/s$,as it penetrates into the wood for a distance of $12\, cm$ is

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