a gun is fired at a 3cm thick solid wooden bar door. the bullet of a mass of 7g travels through the door and has its speed reduced from 450m/s to 175m/s assuming uniform resistance, what is the force of the wood on the bullet?

The force of the wood on the bullet = 20,052 N = 20.052 kN

Explanation:

According to the work-energy theorem, the work done by the wooden bar door in resisting the bullet is equal to the change in kinetic energy of the bullet.

Workdone by the wooden bar door in resisting the bullet = (Resisting force) × (Wood thickness)

Let the constant resisting force be -F (minus sign because it opposes motion)

Wood thickness = 3 cm = 0.03 m

Workdone by the wooden bar door in resisting the bullet = -F × 0.03 = (-0.03F) J

Change in kinetic energy of the bullet = (Final kinetic energy) - (Initial kinetic energy)

Final kinetic energy = ½mv²

m = 7 g = 0.007 kg

v = Final velocity = 175 m/s

Final kinetic energy = ½ × 0.007 × 175² = 107.1875 J

Initial kinetic energy = ½mv²

m = 7 g = 0.007 kg

v = Initial velocity = 450 m/s

Initial kinetic energy = ½ × 0.007 × 450² = 708.75 J

Change in kinetic energy of the bullet as it moves through the wooden bar door = 107.1875 - 708.75 = -601.5625 J

Workdone by the wooden bar door in resisting the bullet = Change in kinetic energy of the bullet as it moves through the wooden bar door

-0.03F = -601.5625

F = (601.5625/0.03) = 20,052.083333333 = 20052 N = 20.052 kN

Hope this Helps!!

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