a bullet as a mass of 0.05kg with an velocity of 80m/s and it enters a wooden block its resistive force act on the bullet and its distance is 0.40m.what force is required to stop the bullet by the wooden block
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According to Newton's second law of motion
Force = mass × velocity / time
here in , final velocity = 0
mass = 0.05kg
velocity = 80m/s
distance = 0.40m
so F = 0.05 × v^2 / 2s ( a = v^2 – u^2/ 2s )
= 0.05 × 6400 / 0.80
= 5 × 6400 / 80
= 5 × 80
= 400N.
Hope it helps!
Force = mass × velocity / time
here in , final velocity = 0
mass = 0.05kg
velocity = 80m/s
distance = 0.40m
so F = 0.05 × v^2 / 2s ( a = v^2 – u^2/ 2s )
= 0.05 × 6400 / 0.80
= 5 × 6400 / 80
= 5 × 80
= 400N.
Hope it helps!
pls mark it best if understood
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Answered by
2
Explanation: as W=DK=F×S=1/2mv^2
So F=1/2×0.05×80×80÷0.4= 400N
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