Lesson Explainer: Newton’s Second Law: Constant Mass | Nagwa Lesson Explainer: Newton’s Second Law: Constant Mass | Nagwa

# Lesson Explainer: Newton’s Second Law: Constant Mass Mathematics

In this explainer, we will learn how to use Newton’s second law of motion with a particle with constant mass under the action of constant force.

Newton’s first law of motion states that if no net force acts on a body, the body does not change in velocity. Newton’s second law of motion states that a net force acting on a particle accelerates the particle.

Let us define Newton’s second law of motion.

### Definition: Newton’s Second Law of Motion

When a net force acts on a body, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body according to the formula where is the net force on the body, is the mass of the body, and is the acceleration of the body.

The weight of a body is an example of a force that acts on the body.

### Definition: The Weight of a Body

The weight of a body is a force that is given by where is the mass of the body and is acceleration due to gravity. If the mass has the unit of kilograms and the acceleration has the unit of metres per second squared, the weight has the unit of newtons. On Earth, the value of is usually taken as 9.8 m/s2.

Let us look at an example of a force acting to accelerate a particle.

### Example 1: Finding a Constant Force Using Newton’s Second Law of Motion

A constant force acted on a body of mass 9 kg such that its speed changed from 58 km/h to 66 km/h in a second. Calculate the magnitude of the force.

The acceleration of the particle is required. Acceleration is usually expressed in units of metres per second squared (m/s2), so it will be necessary to convert the speeds in the question to speeds in metres per second (m/s).

58 km/h in metres per second is given by and 66 km/h in metres per second is given by The increase in speed, , is therefore given by The increase in speed occurs in a time interval of a second, so the acceleration is given by The force that acts to produce this acceleration acts on a mass of 9 kg, and therefore the magnitude of the force is given by

A net force acting in the opposite direction to the direction in which an object is moving acts to decrease the velocity of the object. Let us look at such an example.

### Example 2: Finding the Magnitude of the Force Acting on a Moving Body in the Opposite Direction to Reduce Its Velocity

A body of mass 41 kg was moving along a horizontal road at 14 m/s. A force started acting on the body, opposing its motion. As a result, over the next 26 m, its speed decreased uniformly to 12 m/s. Find the magnitude of the force that caused this change in the body’s motion.

The acceleration of the body is required. Since the distance that the body traveled is stated in the question, we can calculate the acceleration using the kinematic formula where is the speed of the body at the end of the interval, is the speed of the body at the beginning of the interval, and is the displacement of the body along the line of action of the force that acts on it during the interval.

The value of will be negative, as the force acts in the opposite direction to .

To determine the acceleration, the formula must be rearranged to make the subject of the equation:

Substituting the known values gives

The force acting on the body has a magnitude given by

Let us consider Newton’s second law of motion when it is applied to a system where the acceleration is given in terms of vectors.

### Definition: Newton’s Second Law of Motion in terms of Vectors

Given a body of mass with acceleration , the force that is being applied to the body is given by where both and are vector quantities.

We can now look at an example of how this formula can be used.

### Example 3: Finding the Magnitude of a Force Acting on a Body given the Acceleration Vector and Mass of the Body

A body of mass 478 g has an acceleration of m/s2, where and are perpendicular unit vectors. What is the magnitude of the force acting on the body?

First, let us note the units that are used in the question. We have grams (g) and metres per second squared (m/s2). In order to find a solution in terms of N, we will need to convert the mass to kilograms (kg). In doing so, we can say that the mass of the body is 0.478 kg. Now, we can see that our acceleration is given as a vector, so the formula we can use to find the force acting on the body is

Substituting in our values for and , we obtain

This is the vector form of the force acting on the body. The question has asked us to find the magnitude of the force, so we will need to find the magnitude of this vector. We can do this by taking the square root of the sum of the squares of the components as follows:

Hence, our solution is that the magnitude of the force acting on the body is 2.39 N.

Let us look at an example of a body on which forces act in opposite directions.

### Example 4: Finding the Lift Force of a Hot-Air Balloon Accelerating Vertically Downward

A hot-air balloon of mass 1.5 tonnes was accelerating vertically downward at 106.2 cm/s2. Given that the acceleration due to gravity is 9.8 m/s2, find the lift force generated by the hot air.

To simplify the comparison of the vertical forces on the balloon, the acceleration of the balloon can be converted from 106.2 cm/s2 to 1.062 m/s2.

If no vertically upward force acted on the balloon, its vertically downward acceleration would be 9.8 m/s2. The vertically upward lift force on the balloon is the force required to reduce its vertically downward acceleration by

The mass of the balloon is 1.5 tonnes. To obtain a lift force in newtons, this mass must be converted to a mass in kilograms. Since one metric ton is equivalent (equal) to 1‎ ‎000 kg, the mass of the balloon is .

The lift force on the balloon is the force required to accelerate a mass of 1‎ ‎500 kg by 8.738 m/s2.

The magnitude of the force is given by

Let us look at another example of a body on which forces act in opposite directions.

### Example 5: Finding the Speed of a Tank Moving on Resistive Ground

A tank of mass 41 tonnes started moving along a section of horizontal ground. The resistance to its motion was 9 newtons per tonne of its mass, and the magnitude of the force generated by its engine was 1‎ ‎450 N. Determine the tank’s speed 472 seconds after it started moving, rounding the result to the nearest two decimal places.

The resistive force on the tank is 9 newtons per tonne of the mass of the tank. The tank has a mass of 41 tonnes, so the resistive force, , acting on it is given by The force acting to accelerate the tank is 1‎ ‎450 N, so the net force on the tank, , is given by The mass of the tank is 41 tonnes. To determine the acceleration of the tank in metres per second squared, the mass of the tank in kilograms is required, so the mass in tonnes is multiplied by 1‎ ‎000.

The acceleration of the tank is given by The tank accelerates at this rate for 472 s, so the increase in speed of the tank, , is given by To two decimal places, this is given by

The tank is at rest before it starts to move over the resistive ground, so the speed of the tank after 472 seconds is 12.44 m/s.

Now let us look at an example where two bodies accelerate away from each other when the forces acting on the bodies change.

### Example 6: Finding the Distance between an Ascending Balloon and a Body Falling from It given the Balloon’s Velocity

A balloon of mass 1‎ ‎086 kg was ascending vertically at 36 cm/s. If a body of mass 181 kg fell from it, find the distance between the balloon and the body 11 seconds after the body fell. Take .

Initially, the balloon is ascending at a constant speed. The acceleration of the balloon is therefore initially zero. The vertically upward force on the balloon must therefore equal the vertically downward weight of the balloon.

The lift force on the balloon, , has a magnitude equal to that of the initial weight of the balloon, which is given by

When the 181 kg mass is ejected from the balloon, the weight of the balloon decreases, but the lift force on it does not change. The net vertically upward force, , on the balloon is given by

The upward acceleration of the balloon due to is given by The upward displacement of the balloon in 11 seconds due to this acceleration can be found using the kinematic formula and it is given by where is the initial velocity of the balloon. The balloon is initially ascending at 36 cm/s. To make the upward velocity of the balloon consistent with the units of the acceleration , the 36 cm/s is converted to 0.36 m/s.

The value of is therefore given by

In the time that the balloon moves 122.54 m upward, the ejected mass is displaced downward as it falls.

The ejected mass is accelerated downward at 9.8 m/s2 in the opposite direction to , and it initially has the same velocity as the balloon, .

Using the same kinematic equation used to determine , the downward displacement, , is given by

The downward acceleration and displacement are negative as upward displacement has been taken as positive.

The distance, , between the balloon and the ejected mass is given by

### Key Points

• When a net force acts on a body, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body according to the formula where is the mass of the body and is the acceleration of the body.
• For a body on Earth, a weight force acts vertically downward on the body with a magnitude given by where is the mass of the body and .