# 2 questions: Force of bullet and car

• mikex213
In summary, the conversation discusses the calculation of force exerted on a bullet while traveling down a rifle barrel, as well as the resultant of two forces applied to a car and its resulting acceleration. The solution for the first problem involves using the equations d=vt, a=v/t, and f=ma, while the second problem requires drawing a free body diagram to find an idea for solving.
mikex213

## Homework Statement

(1)A 5g bullet leaves the muzzle of a rifle with a speed of v=320m/s. what force (assumed constant) is exerted on the bullet while it is traveling down the 0.82m-long barrel of the rifle?

(2)Two forces are applied to a car in an effort to move it. what is the resultant of these two forces? if the car has a mass of 3000kg, what acceleration does it have? ignore friction.
The car is facing 90degree position, 1st force is 60degrees at 400N, 2nd force is at 100degrees at 450N.

d=vt
a=v/t
f=ma

## The Attempt at a Solution

(1) i tried, d=vt, .82=320m/s*t, t=.003s
a=v/t, a=320m/s / .003s, a=124800m/s^2
f=ma, F= .005kg*124800m/s^2, F=624N

(2) i don't know how to do this one..

for 2) draw the free body diagram first(and post it here) and see if you get an idea.

Welcome to Physics Forums.

mikex213 said:

## The Attempt at a Solution

(1) i tried, d=vt, .82=320m/s*t, t=.003s
The v in that formula should be vaverage. However, you have used vfinal.

What is vaverage, given that vinitial is zero and vfinal is 320 m/s/

a=v/t, a=320m/s / .003s, a=124800m/s^2
f=ma, F= .005kg*124800m/s^2, F=624N
The rest of your work is correct, you just need to get the correct value for t first.

## 1. What is the force of a bullet?

The force of a bullet refers to the amount of energy released when a bullet is fired from a gun. It is typically measured in joules or foot-pounds, and can vary depending on factors such as the caliber of the bullet, the velocity at which it is fired, and the type of gun used.

## 2. How is the force of a bullet calculated?

The force of a bullet can be calculated using the equation F=ma, where F is the force, m is the mass of the bullet, and a is the acceleration it experiences when fired from the gun. This calculation takes into account the muzzle velocity of the bullet and the time it takes for the bullet to travel from the gun to its target.

## 3. What is the force of a car in motion?

The force of a car in motion refers to the amount of energy required to accelerate the car from a state of rest to its current speed. This force is typically measured in Newtons and is influenced by factors such as the mass of the car, the power of the engine, and any external forces acting on the car (such as friction or air resistance).

## 4. How does the force of a car impact its acceleration?

The force of a car directly impacts its acceleration, as described by Newton's Second Law of Motion (F=ma). This means that the greater the force acting on the car, the greater its acceleration will be. However, other factors such as air resistance and rolling resistance can also affect the car's acceleration, so the force alone is not the only determining factor.

## 5. How can the force of a car be increased?

The force of a car can be increased by increasing the power of its engine or by reducing the resistance forces acting against it. This can be done through modifications or enhancements to the car's engine or aerodynamics. However, it is important to note that increasing the force of a car also means an increase in energy consumption and can come with safety risks, so it should be done carefully and responsibly.

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