A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bullet while it is traveling down the barrel of the rifle, 0.820 m long? assume constant acceleration and negligible friction.

Question

Hailee Proctor

Physics

7 April, 13:32

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bullet while it is traveling down the barrel of the rifle, 0.820 m long? assume constant acceleration and negligible friction.

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Answers (1)

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Solomon Vincent · Answer 1

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.

KE = 1/2*m*v^2

= 1/2*0.005*320^2

= 256 Joules.

Divide this by the distance over which this energy was received and you have the force that provided that energy.

= 256/0.820 = 312.195 Newtons

Rounded off, this is 312 N