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

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

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