Newton's Second Law of Motion - Forces and the Laws of Motion - Homework Helpers: Physics

Homework Helpers: Physics

2 Forces and the Laws of Motion

Lesson 2–3: Newton”s Second Law of Motion

One of the implications of the law of inertia is that no force is necessary to keep an object in uniform motion. A spaceship crossing the vast space between galaxies would not need to burn fuel to maintain constant velocity. Its own inertia is all the ship would require to continue moving at a steady speed in the same direction. Expending fuel would be necessary, however, if the pilot of the ship wanted to speed up, slow down, or change directions. Why is that the case? Newton”s second law of motion!

What do you call it when an object speeds up, slows down, or changes direction? If you said a “change in velocity,” you would be correct. What is another word for a change in velocity in a particular period of time? Acceleration. Why would the spaceship need to burn fuel to accelerate? Newton”s first law tells us that in the absence of an unbalanced force, an object in motion will continue in uniform motion. What is the effect of an unbalanced force on the same moving object? An acceleration.

Newton”s second law of motion, the law of acceleration, states that the acceleration an object experiences is directly proportional to the unbalanced force acting on it, and inversely proportional to its own mass. Mathematically, this statement could be expressed with the following formula:

Once again, we can think of many real-life examples that support this law. If you push hard on an object, it moves away from you faster than if you push gently on it. That is because the acceleration that the object experiences is directly proportional (as one goes up, the other goes up) to the force you exert on it. If you use the same amount of force to push a light object as a heavy object, the light object will move away from you with a greater velocity. This is because the acceleration that an object experiences is inversely proportional (as one goes up, the other goes down) to the mass of the object. Both acceleration and force are vector quantities, requiring numbers, units, and direction to describe them completely. It should also be noted that the acceleration that an object experiences is in the same direction as the net force acting on it.

Newton”s second law is usually shown with the symbol for force isolated on one side of the equal sign, as in Fnet = ma or simply F = ma, where the F is assumed to mean “net force.” This seemingly simple formula is very important to your study of physics. Tattooing it onto your body might be a little extreme, but you should become so familiar with it that you come to know it like “the back of your own hand.” It is that important!

Newton”s Second Law
The Law of Acceleration

The acceleration an object experiences is directly proportional to the net unbalanced force exerted upon it and inversely proportional to its mass.

F = ma


In Lesson 2–1 we defined a force as a push or a pull. Newton”s second law shows us that we can also define a force as the product of mass and acceleration. When we multiply the units of mass (kg) by the units of acceleration (m/s2) we get kg · m/s2. Rather than write all of these units every time we describe a force, a derived unit called the newton (N) has been created. It is also important to keep in mind that force is a vector unit, so a direction must be included in a complete description. In many problems, however, no direction will be mentioned. In these cases you are only working with the magnitude (numbers and units) of the force, not direction.

Many of the problems that you encounter in your physics class can be solved with, or at least partially solved with, Newton”s second law. In fact, there is a good chance that there is no other formula that you will use more in physics this year than this one! If your teacher calls on you in class and asks how you should go about solving a motion problem that you don”t know how to start, try saying, “Can we use Newton”s second law of motion?” Chances are, you probably can! Let”s go over some examples of problems that can be solved with this formula.

Example 1

A crate with a mass of 50.0 kg experiences an acceleration with a magnitude of 3.50 m/s2. Find the magnitude of the net force on the crate.

Given: m = 50.0 kg a = 3.50 m/s2

Find: F

Solution: F = ma = (50.0 kg)(3.50 m/s2) = 175 kg × m/s2 = 175 N


That should be easy enough! Remember: As with all of the formulas that you use this year, different variables can be the unknown in a problem. Let”s try an example where acceleration is the unknown.

Example 2

A boy pushes a 0.500 kg book across a table by exerting a net force of 2.00 N on it. Calculate the magnitude of the acceleration that the book will experience.

Given: m = 0.500 kg F = 2.00 N

Find: a


Weight

Now that we have gone over the basic type of problem, let”s go over a couple of special cases. First, you need to understand that what we call “weight” is a specific example of a force. Weight is a measure of the force of gravity between two objects, one of which is usually Earth. If we know the mass of an object, we can determine its weight in newtons by multiplying its mass by its acceleration. Which acceleration do we use? The acceleration due to gravity (g = –9.81 m/s2), provided that we want the weight of the object on or near the surface of the Earth.

Formula for Calculating the Weight of an Object on Earth:
Fw = mg

Understand that this is not a new formula. Rather, it represents a specific case for finding a specific force (force of weight) using a specific acceleration (the acceleration due to gravity). You will be required to use this formula many times this year, so make sure you know when to use it!

Example 3

Find the weight of a person (on Earth) in newtons if they have a mass of 55.0 kg.

Given: m = 55.0 kg g = –9.81 m/s2

Find: Fw

Solution: Fw = mg = (55.0 kg)(–9.81 m/s2) = –539.55 kg · m/s2= -5.40 × 102 N


Do you see why we expressed our answer in scientific notation? I wanted to round to three significant digits, which would force me to round to 540. Using scientific notation allowed me to show three significant digits. Why is the weight in our answer negative? Does that mean that the object is very light? No, the negative sign simply indicates that the direction is downward. Your instructor may have a preference about including the sign in the answer. You will often see weights listed without signs because we are often interested only in the magnitude of the weight, and the direction is understood.

You should keep in mind that the weight of an object is proportional to its mass, but mass and weight are not the same thing. Your weight will change due to location, but your mass won”t. Sometimes a question is designed to test whether or not a student distinguishes between mass and weight. Try Example 4 by yourself, but be careful with it. Some students call this a trick problem, but instructors use this type of problem to test whether or not their students are paying attention to units.

Example 4

A 40.0 N brick experiences a net force 5.00 N parallel to its surface. Find the magnitude of the acceleration that the block will experience.

When the problem refers to a “40.0 N brick” it means that the weight of the brick is 40.0 N. We know that this can”t be the mass of the brick, because the units of mass are kilograms, not newtons. If we want to use Newton”s second law to find the acceleration of the brick, we must first use the special case of Newton”s second law to find the mass of the object.


Now, try the following review problems.

Lesson 2–3 Review

1. _______________ is the force of attraction between two objects due to gravity, where one object is often Earth.

2. What would be the mass of an object that has a weight of 295 N near the surface of Earth?

3. Find the magnitude of the acceleration experienced by a 55.0 kg object with an unbalanced force of 125 N acting on it.

4. Describe a force of “one newton” in words.