# A 4.0 g bullet leaves the muzzle of a rifle with a speed of 326 m/s. What force (assumed...

## Question:

A {eq}4.0\ g {/eq} bullet leaves the muzzle of a rifle with a speed of {eq}326\ \dfrac ms {/eq}. What force (assumed constant) is exerted on the bullet while it is traveling down the {eq}0.79\ m {/eq} long barrel of the rifle?

## Newton's second law of motion

- The second law states that "The force of an object is equal to the product of its mass and acceleration". It is written in the form $$\vec{F} = m\vec{a}$$

The relationship between the force and the mass of the object is directly proportional. This means that an object that is heavier needs a much greater force for it to accelerate. The stronger the force the higher its acceleration.

Acceleration is the objects change in velocity, it is in equation form {eq}a = \frac{\Delta v}{t}{/eq}

where the change in velocity {eq}\Delta v{/eq} is equal to the difference between the final and initial velocity {eq}\Delta v = v_f - v_i{/eq}

## Answer and Explanation:

Given:

- {eq}m = 0.004\ kg{/eq}

- {eq}v_f = 326\ m/s{/eq}

- {eq}x = 0.79\ m{/eq}

To solve this we need to find first the acceleration of the bullet using the formula $$a = \frac{V_f^2}{2x}$$

Substitute the given values$$a = \frac{(326\ m/s)^2}{2(0.79\ m)}$$

$$a = 67 263\ m/s^2$$

Then we use the second law of motion to find the force $$F = ma$$

$$F = (0.004\ kg)( 67 263\ m/s^2)$$

$$\boxed{F = 269\ N}$$

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from GED Science: Life, Physical and Chemical

Chapter 7 / Lesson 6