What will the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a distance of 0.80 m?

The final velocity of the bullet is approximately 3.795 m/s

Explanation:

The given parameters are;

The mass of the bullet, m = 5.0 g

The force applied to the bullet, F = 45 N

The distance over which the force is applied, d = 0.80 m

Therefore, we have;

The final kinetic energy of the bullet = = 1/2 × m × v² = The work done on the bullet by the 45 N force = F × d

Where;

v = The final velocity of the bullet

By substitution, we get;

The final kinetic energy of the bullet = 1/2 × 5.0 × v² = 45 × 0.8

1/2 × 5.0 × v² = 45 × 0.8

v² = (45 × 0.8)/(1/2 × 5.0) = 14.4

v = √14.4 ≈ 3.795

The final velocity of the bullet = v ≈ 3.795 m/s

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