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In essence, Newton's Second Law quantifies the intuitive concept that force is the cause of motion and change. Let's take a look at some examples and the formula to determine how the law of acceleration applies to our everyday lives.
Newton's second law: definition
Below is the definition of Newton's second law, which defines the relationship between the force exerted on an object, its mass, and its acceleration.
Statement: Newton's second law of motion states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.
Newton's second law in action
Mathematically, the formula for Newton's second law is stated as: \begin{equation} Force = mass \cdot acceleration \end{equation} This law is a continuation of Newton's First Law – you may have seen it before without recognising it. Remember that weight is described as \(\text{mass} \cdot \text{gravity}\). We are looking at all these forces being applied to a particle in equilibrium.
So according to the diagram above, we can equate \(\displaystyle F_1 \ + \ F_2 \ + \ F_3\)to 0 because it is in equilibrium (which is when acceleration is 0). But in actual fact, the right-hand side of that equation has always been the\(\mathrm{mass} = 0\).
So far, Newton's First Law applies. However, if the particle starts accelerating, we introduce the value of the acceleration to give us:
\(\displaystyle F_1 \, + \, F_2 \, + \, F_3 \, = \, m \, \cdot \, a\)
\(F_{net} = ma\)
Acceleration is directly proportional to net force and inversely proportional to mass. This implies two things:
Acceleration depends on net force. If the net force is higher, then acceleration will be higher too.
The second quantity that acceleration depends on is the mass of a particle. Let's assume 10 units of force were applied on two balls each with one having a mass of 2kg, and the other of 10kg. The ball with a smaller mass will accelerate more. The smaller the mass, the more the acceleration, and the higher the mass, the lower the acceleration.
SI unit for force
Now we know that force is equal to mass times acceleration, and the SI unit for force is the Newton.
\(\left(kg\right)\left(\frac{m}{s^2}\right) = \frac{kg \cdot m}{s^2}=N\)
Here, mass is measured in kilograms (kg), and acceleration is measured in meters per second squared (\(\textit{m}\textit{s}^{-2}\)). This means you must make sure you have your SI units right when doing calculations.
Sometimes you might have to convert units in order to give your answer in Newtons.
Worked examples of Newton's Second Law
Some clear examples of Newton's Second Law in action are:
Kicking a football: When you kick a football, you apply a force to it. The football accelerates in the direction in which you kicked it. The harder you kick (the greater the force), the faster the ball will accelerate (move).
Car acceleration: When you press the gas pedal in a car, the car moves forward. The harder you press the gas pedal (more force), the faster the car accelerates.
Here are some worked examples of Newton's Second Law. Try and work out some of the problems below by yourself as you read on!
Two people are pushing a car, applying forces of 275N and 395N to the right. Friction provides an opposing force of 560N to the left. If the mass of the car is 1850kg, find its acceleration.
Answer:
Use a bullet point to indicate the car, and place it at the origin of your coordinate system, with y and x. Indicate the forces that are acting on the subject with arrows showing respective direction and magnitude.
First find the total amount of force acting on the body. You will then be able to use that value to find acceleration.
\(\displaystyle F_{net} = m \cdot a\)
275 + 395 -560 = 1850a
560 here is a negative value because it clearly states in the question that is an opposing force. This is also why it is shown in the negative direction on our diagram.
110 = 1850a
Divide both sides by 1850 to find the acceleration.
\begin{equation*} a \, = \, \frac{110}{1850} \end{equation*}
\(a\phantom{ }\!=\phantom{ }\!0.059ms^{-2}\)
The car is accelerating at \(\displaystyle a\ =\ 0.059\,m\,s^{-2}\)
You have an 8kg block and you apply a force of 35N west. The block is on a surface that opposes it with a force of 19N.
Calculate the net force.
Calculate the direction of the acceleration factor.
Answer: You may want to draw your diagram to help visualise the situation.
35N is acting in the negative direction, and 19N is acting in the positive direction. So finding the net force will be carried out like this:
\(\displaystyle F_{net} = 19N - 35N\)
\(\textstyle F_{net} = -16N\)
Net force here is -16 N .
If you are asked to find the magnitude of the force, your answer should be a positive figure because the magnitude of a vector is always positive. The negative sign tells you the direction of the force. So the magnitude of the force in this example is 16N.
Once you find the net force, you can find the acceleration.
\(F_{net} = ma\)
-16 = 8a
\(\displaystyle a \ = \ -2ms^{-2}\)
The negative value here tells us that the acceleration is towards the left. Therefore, the block is slowing down.
Newton's second law and inclined planes
An inclined plane is a sloping surface over which loads can be lowered or raised. The rate at which a particle accelerates on an inclined plane is very significant to its degree of slope. This means that the bigger the slope, the bigger the acceleration will be on the particle.
If a particle of mass 2kg is released from rest on a smooth slope inclined to the horizontal at an angle of 20°, what will the acceleration of the block be?
A smooth slope (or similar wording) tells you that there is no friction involved.
Answer: Model this graphically to help with the calculation.
This diagram (or a similar one) could be given to you in the question. However, you can modify the diagram to understand it better. Draw an x and y-axis perpendicular to the inclined particle to help you determine which forces are working on your particle.
As you can see, the only significant force acting on the particle is gravity. There is also a 20° angle between the vertical force and the displaced perpendicular line to the particle. That is obviously 20° because of the degree of slope. If the plane slopes at 20°, the displaced angle will also be 20°.
Since we are looking for acceleration, we will focus on the forces parallel to the plane.
\(\begin{equation*} F_{net} = ma \end{equation*}\)
We will now split the force into vertical and horizontal opponents using trigonometry.
\(\text{sin}\:\theta=\frac{\text{Opposite Side}}{\text{Hypotenuse}}\)
\(\text{Opposite side} = \text{Hypotenuse} \cdot \sin{\theta}\)
2g sin20 = 2a
a = g sin20
\(\displaystyle a \ = \ 3 \cdot 4ms^{-2}\)
Newton's Second Law - Key takeaways
- Your force can only be in Newtons when you have your mass measured in kilograms (kg), and your acceleration in meters per second
\(\left(m s^{-2}\right)\) - Newton's second law of motion states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.
- Newton's second law of motion is mathematically written as \(\text{Force} = \text{mass} \cdot \text{acceleration}\).
- An inclined plane is a sloping surface over which loads can be lowered or raised.
- The higher the degree of the slope in an inclined plane, the more acceleration a particle will possess.
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Frequently Asked Questions about Newton's Second Law
Why is Newton's second law important?
Newton's second law shows us the relationship between forces and motion.
What is the definition of Newton's second law?
Newton's second law of motion states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.
Does Newton's second law apply to rockets?
Yes
What is the equation for Newton's second law of motion?
Fnet = ma
How does Newton's second law apply to a car crash?
The force a car possesses increases when either the acceleration or mass is increased. This means that a car that weighs 900kg will possess more force in a crash than one that weighs 500kg if the acceleration in both were the same.
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