In physics terms, acceleration, a, is the amount by which your velocity changes in a given amount of time. Given the initial and final velocities, vi and vf, and the initial and final times over which your speed changes, ti and tf, you can write the equation like this:

In terms of units, the equation looks like this:

Distance per time squared? Don’t let that throw you. You end up with time squared in the denominator because you divide velocity by time. In other words, acceleration is the rate at which your velocity changes, because rates have time in the denominator. For acceleration, you see units of meters per second 2, centimeters per second 2, miles per second 2, feet per second 2, or even kilometers per hour 2.

It may be easier, for a given problem, to use units such as mph/s (miles per hour per second). This would be useful if the velocity in question had a magnitude of something like several miles per hour that changed typically over a number of seconds.

Say you become a drag racer in order to analyze your acceleration down the dragway. After a test race, you know the distance you went — 402 meters, or about 0.25 miles (the magnitude of your displacement) — and you know the time it took — 5.5 seconds. So what was your acceleration as you blasted down the track?

Well, you can relate displacement, acceleration, and time as follows:

and that’s what you want — you always work the algebra so that you end up relating all the quantities you know to the one quantity you don’t know. In this case, you have

(Keep in mind that in this case, your initial velocity is 0 — you’re not allowed to take a running start at the drag race!) You can rearrange this equation with a little algebra to solve for acceleration; just divide both sides by t2 and multiply by 2 to get

Great. Plugging in the numbers, you get the following:

Okay, the acceleration is approximately 27 meters per second 2. What’s that in more understandable terms? The acceleration due to gravity, g, is 9.8 meters per second 2, so this is about 2.7 g’s — you’d feel yourself pushed back into your seat with a force about 2.7 times your own weight.

## Hooke’s law

Bourdon tubes are based on Hooke’s law. The force created by gas pressure inside the coiled metal tube above unwinds it by an amount proportional to the pressure.

The balance wheel at the core of many mechanical clocks and watches depends on Hooke’s law. Since the torque generated by the coiled spring is proportional to the angle turned by the wheel, its oscillations have a nearly constant period.

Hooke’s law is a law of physics that states that the force (F) needed to extend or compress a spring by some distance x scales linearly with respect to that distance.

That is.** displaystyle Fs=kx displaystyle Fs=kx**, where k is a constant factor characteristic of the spring: its stiffness, and x is small compared to the total possible deformation of the spring. The law is named after 17th-century British physicist Robert Hooke. He first stated the law in 1676 as a Latin anagram. He published the solution of his anagram in 1678 as: ut tensio, sic vis (“as the extension, so the force” or “the extension is proportional to the force”). Hooke states in the 1678 work that he was aware of the law already in 1660.

Hooke’s equation holds (to some extent) in many other situations where an elastic body is deformed, such as wind blowing on a tall building, a musician plucking a string of a guitar, and the filling of a party balloon. An elastic body or material for which this equation can be assumed is said to be linear-elastic or Hookean.

Hooke’s law is only a first-order linear approximation to the real response of springs and other elastic bodies to applied forces. It must eventually fail once the forces exceed some limit, since no material can be compressed beyond a certain minimum size, or stretched beyond a maximum size, without some permanent deformation or change of state. Many materials will noticeably deviate from Hooke’s law well before those elastic limits are reached.

On the other hand, Hooke’s law is an accurate approximation for most solid bodies, as long as the forces and deformations are small enough. For this reason, Hooke’s law is extensively used in all branches of science and engineering, and is the foundation of many disciplines such as seismology, molecular mechanics and acoustics. It is also the fundamental principle behind the spring scale, the manometer, and the balance wheel of the mechanical clock.

The modern theory of elasticity generalizes Hooke’s law to say that the strain (deformation) of an elastic object or material is proportional to the stress applied to it. However, since general stresses and strains may have multiple independent components, the “proportionality factor” may no longer be just a single real number, but rather a linear map (a tensor) that can be represented by a matrix of real numbers.

In this general form, Hooke’s law makes it possible to deduce the relation between strain and stress for complex objects in terms of intrinsic properties of the materials it is made of. For example, one can deduce that a homogeneous rod with uniform cross section will behave like a simple spring when stretched, with a stiffness k directly proportional to its cross-section area and inversely proportional to its length.

## HOW TO CALCULATE A SPRING CONSTANT USING HOOKE’S LAW

Any physicist knows that if an object applies a force to a spring, then the spring applies an equal and opposite force to the object. Hooke’s law gives the force a spring exerts on an object attached to it with the following equation:

the minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. The variables of the equation are: F which represents force, k which is called the spring constant and measures how stiff and strong the spring is, and x is the distance the spring is stretched or compressed away from its equilibrium or rest position.

The force exerted by a spring is called a restoring force; it always acts to restore the spring toward equilibrium. In Hooke’s law, the negative sign on the spring’s force means that the force exerted by the spring opposes the spring’s displacement.

### UNDERSTANDING SPRINGS AND THEIR DIRECTION OF FORCE

direction of force exerted in springs

The direction of force exerted by a spring.

The preceding figure shows a ball attached to a spring. You can see that if the spring isn’t stretched or compressed, it exerts no force on the ball. If you push the spring, however, it pushes back, and if you pull the spring, it pulls back.

Hooke’s law is valid as long as the elastic material you’re dealing with stays elastic — that is, it stays within its elastic limit. If you pull a spring too far, it loses its stretchy ability. As long as a spring stays within its elastic limit, you can say that F = –kx. When a spring stays within its elastic limit and obeys Hooke’s law, the spring is called an ideal spring.

### HOW TO FIND THE SPRING CONSTANT (EXAMPLE PROBLEM)

Suppose that a group of car designers knocks on your door and asks whether you can help design a suspension system. “Sure,” you say. They inform you that the car will have a mass of 1,000 kilograms, and you have four shock absorbers, each 0.5 meters long, to work with. How strong do the springs have to be? Assuming these shock absorbers use springs, each one has to support a mass of at least 250 kilograms, which weighs the following:

F = mg = (250 kg)(9.8 m/s2) = 2,450 N

where F equals force, m equals the mass of the object, and g equals the acceleration due to gravity, 9.8 meters per second2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke’s law says:

F = –kx

Now, we need to rework the equation so that we are calculating for the missing metric which is the spring constant, or k. Looking only at the magnitudes and therefore omitting the negative sign, you get

Time to plug in the numbers:

The springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter. The car designers rush out, ecstatic, but you call after them, “Don’t forget, you need to at least double that if you actually want your car to be able to handle potholes.”

### HOW TO CALCULATE ANGULAR MOMENTUM

A small child on a spinning playground ride, such as a merry go round, and she’s yelling that she wants to get off. You have to stop the spinning ride, but it’s going to take some effort. Why? Because it has *angular momentum.*

In physics, you can calculate angular momentum in the same way that you calculate linear momentum just substitute moment of inertia for mass, and angular velocity for velocity.

### WHAT IS ANGULAR MOMENTUM?

**Angular momentum.**

The quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity.

*Linear momentum,* **p,** is defined as the product of mass and velocity:

#### p = *m*v

This is a quantity that is conserved when there are no external forces acting. The more massive and faster moving an object, the greater the magnitude of momentum.

### THE ANGULAR MOMENTUM EQUATION

Physics also features angular momentum, **L.** The equation for angular momentum looks like this:

The angular momentum equation features three variables:

- L = angular momentum
- / = the moment of inertia
- W = the angular velocity

Note that angular momentum is a vector quantity, meaning it has a magnitude and a direction.

the thumb of your right hand points when you wrap your fingers around in the direction the object is turning).

in the MKS (meter-kilogram-second) system.

The important idea about angular momentum, much as with linear momentum, is that it’s conserved.

The *principle of conservation of angular momentum *states that angular momentum is conserved if no net torques are involved.

This principle comes in handy in all sorts of problems, such as when two ice skaters start off holding each other close while spinning but then end up at arm’s length. Given their initial angular velocity, you can find their final angular velocity, because angular momentum is conserved:

If you can find the initial moment of inertia and the final moment of inertia, you’re set. But you also come across less obvious cases where the principle of conservation of angular momentum helps out. For example, satellites don’t have to travel in circular orbits; they can travel in ellipses. And when they do, the math can get a lot more complicated. Lucky for you, the principle of conservation of angular momentum can make the problems simple.

### ANGULAR MOMENTUM EXAMPLE PROBLEM

Say that NASA planned to put a satellite into a circular orbit around Pluto for studies, but the situation got a little out of hand and the satellite ended up with an elliptical orbit. At its nearest point to Pluto,

the satellite zips along at 9,000 meters per second.

at that point? The answer is tough to figure out unless you can come up with an angle here, and that angle is angular momentum.

Angular momentum is conserved because there are no external torques the satellite must deal with (gravity always acts parallel to the orbital radius). Because angular momentum is conserved, you can say that

Because the satellite is so small compared to the radius of its orbit at any location, you can consider the satellite a point mass. Therefore, the moment of inertia, *I,* equals *mr*^{2}. The magnitude of the angular velocity equals *v/r*, so you can express the conservation of angular momentum in terms of the velocity like so:

You can put *v*_{2}on one side of the equation by dividing by *mr*_{2}*:*

You have your solution; no fancy math involved at all, because you can rely on the principle of conservation of angular momentum to do the work for you. All you need to do is plug in the numbers:

At its closest point to Pluto, the satellite will be screaming along at 9,000 meters per second, and at its farthest point, it will be moving at 2,700 meters per second. Easy enough to figure out, as long as you have the principle of conservation of angular momentum under your belt.

### HOW TO CALCULATE DISPLACEMENT IN A PHYSICS PROBLEM

Displacement is the distance between an object’s initial position and its final position and is usually measured or defined along a straight line. Since this is a calculation that measures distance, the standard unit is the meter (m).

### HOW TO FIND DISPLACEMENT

In physics, you find displacement by calculating the distance between an object’s initial position and its final position.

In physics terms, you often see displacement referred to as the variable s.

The official displacement formula is as follows:

s = sf – si

s = displacement

si = initial position

sf = final position

#### CALCULATING DISPLACEMENT EXAMPLE

Say, for example, that you have a fine new golf ball that’s prone to rolling around. This particular golf ball likes to roll around on top of a large measuring stick and you want to know how to calculate displacement when the ball moves. You place the golf ball at the 0 position on the measuring stick, as shown in the below figure, diagram A.

Examining displacement with a golf ball.

The golf ball rolls over to a new point, 3 meters to the right, as you see in the figure, diagram B. The golf ball has moved, so displacement has taken place. In this case, the displacement is just 3 meters to the right. Its initial position was 0 meters, and its final position is at +3 meters. The displacement is 3 meters.

Scientists, being who they are, like to go into even more detail. You often see the term si, which describes initial position, (the i stands for initial). And you may see the term sf used to describe final position.

In these terms, moving from diagram A to diagram B in the figure, si is at the 0-meter mark and sf is at +3 meters. The displacement, s, equals the final position minus the initial position

Displacements don’t have to be positive; they can be zero or negative as well. If the positive direction is to the right, then a negative displacement means that the object has moved to the left.

In diagram C, the restless golf ball has moved to a new location, which is measured as –4 meters on the measuring stick. The displacement is given by the difference between the initial and final position. If you want to know the displacement of the ball from its position in diagram B, take the initial position of the ball to be si = 3 meters; then the displacement is given by

When working on physics problems, you can choose to place the origin of your position-measuring system wherever is convenient. The measurement of the position of an object depends on where you choose to place your origin; however, displacement from an initial position si to a final position sf does not depend on the position of the origin because the displacement depends only on the difference between the positions, not the positions themselves.

### HOW TO CALCULATE ACCELERATION

In physics terms, acceleration, a, is the amount by which your velocity changes in a given amount of time. Given the initial and final velocities, vi and vf, and the initial and final times over which your speed changes, ti and tf, you can write the equation like this:

In terms of units, the equation looks like this:

Distance per time squared? Don’t let that throw you. You end up with time squared in the denominator because you divide velocity by time. In other words, acceleration is the rate at which your velocity changes, because rates have time in the denominator. For acceleration, you see units of meters per second 2, centimeters per second 2, miles per second 2, feet per second 2, or even kilometers per hour 2.

It may be easier, for a given problem, to use units such as mph/s (miles per hour per second). This would be useful if the velocity in question had a magnitude of something like several miles per hour that changed typically over a number of seconds.

Say you become a drag racer in order to analyze your acceleration down the dragway. After a test race, you know the distance you went — 402 meters, or about 0.25 miles (the magnitude of your displacement) — and you know the time it took — 5.5 seconds. So what was your acceleration as you blasted down the track?

Well, you can relate displacement, acceleration, and time as follows:

and that’s what you want — you always work the algebra so that you end up relating all the quantities you know to the one quantity you don’t know. In this case, you have

(Keep in mind that in this case, your initial velocity is 0 — you’re not allowed to take a running start at the drag race!) You can rearrange this equation with a little algebra to solve for acceleration; just divide both sides by t2 and multiply by 2 to get

Great. Plugging in the numbers, you get the following:

Okay, the acceleration is approximately 27 meters per second 2. What’s that in more understandable terms? The acceleration due to gravity, g, is 9.8 meters per second 2, so this is about 2.7 g’s — you’d feel yourself pushed back into your seat with a force about 2.7 times your own weight.

### HOW TO FIND A VECTOR’S MAGNITUDE AND DIRECTION

If you’re given the components of a vector, such as (3, 4), you can convert it easily to the magnitude/angle way of expressing vectors using trigonometry.

For example, take a look at the vector in the image.

Suppose that you’re given the coordinates of the end of the vector and want to find its magnitude, v, and angle, theta. Because of your knowledge of trigonometry, you know

Where tan theta is the tangent of the angle. This means that

theta = tan–1(y/x)

Suppose that the coordinates of the vector are (3, 4). You can find the angle theta as the tan–1(4/3) = 53 degrees.

You can use the Pythagorean theorem to find the hypotenuse — the magnitude, v — of the triangle formed by x, y, and v:

Plug in the numbers for this example to get

So if you have a vector given by the coordinates (3, 4), its magnitude is 5, and its angle is 53 degrees.

### SAMPLE QUESTION

- Convert the vector given by the coordinates (1.0, 5.0) into magnitude/angle format.

- The correct answer is magnitude 5.1, angle 79 degrees.

- Apply the Pythagorean theorem to find the magnitude. Plug in the numbers to get 5.1.

- Apply the equation theta= tan–1(y/x) to find the angle. Plug in the numbers to get tan–1(5.0/1.0) = 79 degrees.

### PRACTICE QUESTIONS

- Convert the vector (5.0, 7.0) into magnitude/angle form.

- Convert the vector (13.0, 13.0) into magnitude/angle form.

- Convert the vector (–1.0, 1.0) into magnitude/angle form.

- Convert the vector (–5.0, –7.0) into magnitude/angle form.

### Following are answers to the practice questions:

Magnitude 8.6, angle 54 degrees

Apply the equation

to find the magnitude, which is 8.6.

Apply the equation theta = tan–1(y/x) to find the angle: tan–1(7.0/5.0) = 54 degrees.

Magnitude 18.4, angle 45 degrees

Apply the equation

to find the magnitude, which is 18.4.

Apply the equation theta = tan–1(y/x) to find the angle: tan–1(13.0/13.0) = 45 degrees.

Magnitude 1.4, angle 135 degrees

Apply the equation

to find the magnitude, which is 1.4.

Apply the equation theta = tan–1(y/x) to find the angle: tan–1(1.0/–1.0) = –45 degrees.

However, note that the angle must really be between 90 degrees and 180 degrees because the first vector component is negative and the second is positive. That means you should add 180 degrees to –45 degrees, giving you 135 degrees (the tangent of 135 degrees is also 1.0/–1.0 = –1.0).

Magnitude 8.6, angle 234 degrees

Apply the equation

to find the magnitude, which is 8.6.

Apply the equation theta = tan–1(y/x) to find the angle: tan–1(–7.0/–5.0) = 54 degrees.

However, note that the angle must really be between 180 degrees and 270 degrees because both vector components are negative. That means you should add 180 degrees to 54 degrees, giving you 234 degrees (the tangent of 234 degrees is also –7.0/–5.0 = 7.0/5.0).

### IMPORTANT PHYSICS EQUATIONS TO REMEMBER

Physics is packed with formulas and equations. This comprehensive list, arranged by topic, represents essential physics equations you need to keep handy when you’re dealing with physics formulas.

### HOW TO CALCULATE TIME AND DISTANCE FROM ACCELERATION AND VELOCITY

In a physics equation, given a constant acceleration and the change in velocity of an object, you can figure out both the time involved and the distance traveled. For instance, imagine you’re a drag racer. Your acceleration is 26.6 meters per second 2, and your final speed is 146.3 meters per second. Now find the total distance traveled. Got you, huh? “Not at all,” you say, supremely confident. “Just let me get my calculator.”

You know the acceleration and the final speed, and you want to know the total distance required to get to that speed. This problem looks like a puzzler, but if you need the time, you can always solve for it. You know the final speed, vf, and the initial speed, vi (which is zero), and you know the acceleration, a. Because vf – vi = at, you know that

Now you have the time. You still need the distance, and you can get it this way:

The second term drops out because vi = 0, so all you have to do is plug in the numbers:

In other words, the total distance traveled is 402 meters, or a quarter mile. Must be a quarter-mile racetrack.

### IMPORTANT PHYSICS EQUATIONS TO REMEMBER

Many people find Physics to be a difficult subject to approach. Well now, you have some tools to help you along the way. This handy list of physics equations organized by topic contains the most common equations you’ll run across.

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