Physics, asked by Anonymous, 4 months ago

A bullet of mass 30 g moving with a speed of 500 ms penetrates 10 cm into a fixed target. Calculate the average force exerted by target on the bullet.

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Answers

Answered by Anonymous
9

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Question

A bullet of mass 30 g moving with a speed of 500 ms penetrates 10 cm into a fixed target. Calculate the average force exerted by target on the bullet.

Solution

Mass of bullet,

m = \frac{30}{1000} g = 0.03kg

Speed (v) =  {500 \: ms}^{ - 1}

Initial kinetic energy =  \frac{1}{2}  {mv}^{2}

 =  \frac{1}{2}  \times 0.03 \times ( {500})^{2}

 =  \frac{1}{2}  \times 0.03 \times 250000 = 3750J

Final Kinetic Energy  =  \frac{1}{2} {mv}^{2} = 0

Loss in kinetic energy = 3750 J

Suppose:

F = average force applied by block on bullet

S = displacement = 10 cm = 0.10m

Applying work energy principle,

W = Δk

Fs = Δk

F × 10 = 3750

F \:  =  \:  \frac{3750}{0.10}  = 3.5 \times  {10}^{4} N

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Answered by jaya8025
0

Elllooooo Sry refer the above one for your answer kindly

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