Lesson Explainer: Newton’s Second Law of Motion Physics • 9th Grade

In this explainer, we will learn how to apply Newton’s second law of motion, 𝐹=π‘šπ‘Ž, to find the accelerations caused by forces that act in various directions.

Newton’s first law of motion explains that an object does not change velocity unless a net force acts on it and that the velocity of the object must change if a net force acts on it. Newton’s second law of motion explains how much the velocity of an object will change when a net force acts on it.

Newton’s second law of motion tells us that the change in the velocity of an object depends on the force applied to it. For an object to change velocity, the object must accelerate. The acceleration of an object is the rate at which the velocity of the object changes as time passes. Newton’s second law of motion states that the acceleration of an object is directly proportional to the force applied to the object. This can be expressed as πΉβˆπ‘Ž, where 𝐹 is the force applied and π‘Ž is the acceleration of the object.

Both force and acceleration are vector quantities; hence, the acceleration of an object is in the direction of the force applied to it.

Newton’ second law of motion is usually expressed as 𝐹=π‘šπ‘Ž, where π‘š is the constant of proportionality between the force applied to an object and the acceleration of the object that results. This is called the mass of the object. The greater the mass of an object, the greater the force required to accelerate the object.

The mass of an object is also related to the gravitational force produced by the object, but in this explainer, we are only concerned with mass as the constant of proportionality between force and acceleration. This is termed the inertial mass. All mention of mass in this explainer refers to inertial mass.

Let us look at an example in which a force acts on an object that changes velocity.

Example 1: Determining the Force Acting on an Object that Changes Velocity

An object with a mass of 22 kg has a force applied to it. The graph shows the change in the object’s velocity while the force is applied. How much force is applied to the object? Answer to the nearest newton.

Answer

The force on the object is given by the formula 𝐹=π‘šπ‘Ž, where the mass of the object, π‘š, is 22 kg and the acceleration, π‘Ž, is not given.

The acceleration of the object is the rate of change of the velocity as time passes. The graph shows the change in the velocity of the object with time.

The graph shows that the velocity of the object changes from 5 m/s to 20 m/s. The increase in velocity is therefore given by Δ𝑣=20/βˆ’5/=15/.msmsms

The graph shows that the time taken for the object to increase in velocity is 5 seconds.

The acceleration of the object is the rate of change of the velocity as time passes, so it is given by π‘Ž=15/5=3/.mssms

Now that the acceleration and the mass of the object are known, the force on it can be determined as 𝐹=22Γ—3/=66.kgmsN

Let us now look at an example in which the relationship between force and acceleration is stated more directly.

Example 2: Determining the Force Acting on an Accelerating Particle

How much force is applied to an object of mass 5 kg that is accelerated by that force at a rate of 2 m/s2?

Answer

The force on the object is given by the formula 𝐹=π‘šπ‘Ž, where the mass of the object, π‘š, is 5 kg and the acceleration, π‘Ž, is 2 m/s2.

We have then that 𝐹=5Γ—2/=10.kgmsN

Let us now look at an example in which the force acting on a real-world object is determined.

Example 3: Determining the Force Applied to Accelerate a System of Objects

A skateboarder of mass 50 kg stands on a skateboard of mass 1.5 kg, close to the wall of a building. The skateboarder stands with one foot on the board and uses the other foot to push away from the wall. The skateboard accelerates away from the wall at 4.66 m/s2. With how much force did the skateboarder push against the wall? Give your answer to the nearest newton.

Answer

The force on the object is given by the formula 𝐹=π‘šπ‘Ž, where π‘š is the mass of the object that the force acts on and the acceleration, π‘Ž, is 4.66 m/s2.

The skateboarder stands on the skateboard. When the skateboard accelerates away from the wall, the skateboarder is also accelerated away from the wall at the same rate (otherwise, the question would presumably mention something about the skateboarder falling off the skateboard). The mass that is accelerated is then the mass of the skateboard and the skateboarder. This mass is given by π‘š=1.5+50=51.5.kgkgkg

We have then that 𝐹=51.5Γ—4.66/=239.99.kgmsN

The question asks for the force to the nearest newton, which is 240 N.

Newton’s second law of motion, expressed as the formula 𝐹=π‘šπ‘Ž, describes the relationship between the force acting on an object, the mass of the object, and the acceleration of the object, and so any of these quantities can be determined using the formula. To find the mass or acceleration of an object using the formula for Newton’s second law of motion, the formula must be rearranged to change the subject of the formula to the quantity being determined.

For example, to determine the acceleration of the object using this formula, π‘Ž must be made the subject of the formula. We can do this by first dividing the formula by π‘š: πΉπ‘š=π‘šπ‘Žπ‘š.

The right side of the resulting equation has the term π‘š in both the numerator and the denominator: πΉπ‘š=ο€»π‘šπ‘šο‡π‘Ž.

Dividing π‘š by π‘š gives the result 1: πΉπ‘š=1π‘Ž.1π‘Ž equals π‘Ž, so the expression for π‘Ž is given by π‘Ž=πΉπ‘š.

Let us now look at an example in which the acceleration of an object is determined using Newton’s second law of motion.

Example 4: Determining the Acceleration of an Object on Which a Force Acts

An object of mass 1.5 kg has a force of 4.5 N applied to it. At what rate does the force accelerate the object?

Answer

Newton’s second law of motion can be expressed by the formula 𝐹=π‘šπ‘Ž, where 𝐹 the is force that acts on the object, π‘š is the mass of the object, and π‘Ž is the acceleration.

To determine the acceleration of the object using this formula, π‘Ž must be made the subject of the formula. We can do this by dividing the formula by π‘š: πΉπ‘š=π‘šπ‘Žπ‘š=π‘Žπ‘Ž=πΉπ‘š.

Substituting the values in the question, we see that π‘Ž=4.51.5=3/.Nkgms

Let us now look at an example in which the mass of an object is determined using Newton’s second law of motion.

Example 5: Determining the Mass of an Object on Which a Force Acts

An object accelerates at 4 m/s2 while a force of 20 N is applied to it. What is the mass of the object?

Answer

Newton’s second law of motion can be expressed by the formula 𝐹=π‘šπ‘Ž, where 𝐹 the is force that acts on the object, π‘š is the mass of the object, and π‘Ž is the acceleration.

To determine the mass of the object using this formula, π‘š must be made the subject of the formula. We can do this by dividing the formula by π‘Ž: πΉπ‘Ž=π‘šπ‘Žπ‘Ž=π‘šπ‘š=πΉπ‘Ž.

Substituting the values in the question, we see that π‘š=204/=5.Nmskg

Let us now look at an example involving determining the acceleration of a real-world object.

Example 6: Determining the Acceleration of an Object on Which a Force Acts

A ball has a mass of 250 g. The ball is kicked and this applies a force of 15 N to the ball, as shown in the diagram. How much does the ball accelerate by in the direction of the kick?

Answer

Newton’s second law of motion can be expressed by the formula 𝐹=π‘šπ‘Ž, where 𝐹 the is force that acts on the object, π‘š is the mass of the object, and π‘Ž is the acceleration.

To determine the acceleration of the object using this formula, π‘Ž must be made the subject of the formula. We can do this by dividing the formula by π‘š: πΉπ‘š=π‘šπ‘Žπ‘š=π‘Žπ‘Ž=πΉπ‘š.

The mass of the ball is 250 grams, but the formula 𝐹=π‘šπ‘Ž gives the force in newtons acting on a mass in kilograms required to accelerate an object by a number of metres per second squared.

The force on the ball is given in newtons, so to determine the acceleration of the ball in metres per second squared, the mass of the ball must be given in kilograms. A mass of 250 grams is equivalent to a mass of 0.250 kilograms.

The acceleration of the ball in the direction of the kick is given by π‘Ž=πΉπ‘šπ‘Ž=150.250=60/.Nkgms

Realistically, this does not mean that the velocity of the ball after it is kicked is 50 m/s.

This would be true only if the boot that kicked the ball kicked it with an average force of 15 newtons throughout 1 second and if no other horizontal force acted on the ball while it was kicked.

When kicking a ball, the contact between the kicking foot and the ball persists for much less than 1 second. Also, a ball in contact with the ground is acted on by friction, which reduces its velocity, as does air resistance.

Force and acceleration are both vector quantities and so have direction as well as magnitude. When only one force acts on an object, the direction of the acceleration of the object is necessarily in the direction of the force.

Multiple forces can act on an object, and these forces do not necessarily act in the same direction. The acceleration of an object that multiple forces act on depends on the net force on the object.

Let us look at an example in which multiple forces act on an object.

Example 7: Determining the Mass of an Object on Which Two Forces Act

An object has forces of 30 N and 55 N applied to it. The forces act in opposite directions to each other, as shown in the diagram. The object accelerates to the left at 0.5 m/s2. What is the mass of the object?

Answer

Newton’s second law of motion can be expressed by the formula 𝐹=π‘šπ‘Ž, where 𝐹 is the force that acts on the object, π‘š is the mass of the object, and π‘Ž is the acceleration.

In this case, the force that acts on the object is the resultant of two forces that act in opposite directions: a 30 N force acting to the right and a 55 N force acting to the left.

The object is stated to accelerate to the left, which is in the same direction as the 55 N force. The direction of acceleration is in the direction of the greater force.

Taking left as positive, a force acting to the left has a positive magnitude and a force acting to the right has a negative magnitude.

The magnitudes of the forces acting on the object are, therefore, 55 N and –30 N. The sum of these forces is given by 𝐹=55+(βˆ’30)=55βˆ’30=25.NNNNN

Now that the force has been determined, to determine the mass of the object using the formula 𝐹=π‘šπ‘Ž,π‘š must be made the subject of the formula. We can do this by dividing the formula by π‘Ž: πΉπ‘Ž=π‘šπ‘Žπ‘Ž=π‘šπ‘š=πΉπ‘Ž.

Substituting the values in the question, we see that π‘š=250.5/=50.Nmskg

Let us now look at an example involving multiple forces acting on real-world objects.

Example 8: Determining the Acceleration of an Object on Which Two Forces Act

A swimmer of mass 48 kg uses her legs to push herself away from a swimming pool wall, applying a force of 280 N. The water that the swimmer accelerates through applies 160 N of force in the opposite direction to that in which she accelerates. What is the swimmer’s acceleration through the water?

Answer

Newton’s second law of motion can be expressed by the formula 𝐹=π‘šπ‘Ž, where 𝐹 the is force that acts on the object, π‘š is the mass of the object, and π‘Ž is the acceleration.

In this case, the force that acts on the object is the resultant of two forces that act in opposite directions: a 280 N force acting away from the pool wall and a 160 N force acting toward the pool wall.

Taking the direction away from the pool wall as positive, a force acting away from the pool wall has a positive magnitude and a force acting toward the pool wall has a negative magnitude.

The magnitudes of the forces acting on the object are, therefore, 280 N and –160 N. The sum of these forces is given by 𝐹=280+(βˆ’160)=280βˆ’160=120.NNNNN

Now that the force has been determined, to determine the acceleration of the swimmer using the formula 𝐹=π‘šπ‘Ž,π‘Ž must be made the subject of the formula. We can do this by dividing the formula by π‘š: πΉπ‘Ž=π‘šπ‘Žπ‘š=π‘Žπ‘Ž=πΉπ‘š.

Substituting the values in the question, we see that π‘Ž=12048=2.5/.Nkgms

It is fairly realistic that the swimmer’s legs could be in contact with the pool wall for one second before the swimmer’s feet lose contact with the pool wall, if the swimmer’s legs were initially bent and they straightened over a time of one second. The force acting in the opposite direction to the swimmer’s motion has been included in the calculation, so the result of 2.5 m/s2 in this case could reasonably mean that the swimmer has a velocity of 2.5 m/s at the instant that her feet lose contact with the pool wall.

This velocity would not be maintained, however, as the resistance of the water would continue to act on her, and without contact between her feet and the pool wall, the only force acting on her would be this resistance; hence, her velocity would decrease.

Let us now summarize what we have learned in these examples.

Key Points

  • The acceleration of an object that a force acts on is directly proportional to the force acting on and to the mass of the object. This relationship can be expressed by the formula 𝐹=π‘šπ‘Ž, where 𝐹 is the force that acts on the object, π‘š is the mass of the object, and π‘Ž is the acceleration of the object.
  • The formula 𝐹=π‘šπ‘Ž can be rearranged to determine the mass or the acceleration of an object.
  • When multiple forces act on an object, the acceleration of the object is directly proportional to the resultant of these forces.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.