**Newton’s second law equation**, a = F_{net}/m, establishes the relationship between the acceleration (a) of an object, the net force (F_{net}) acting on it, and its mass (m). This equation states that the acceleration of an object is directly proportional to the applied net force and inversely proportional to its mass. By utilizing this equation, it becomes possible to quantitatively analyze and predict the acceleration of an object when the values of net force and mass are known.

## Practice problems

### Problem #1

A box with a mass of 2 kg accelerates forward slowly when a net force of 4 N is applied to it. What is the rate of acceleration of the box?

**Solution**

Given data:

- Mass of a box, m = 2 kg
- Net force applied to a box, F
_{net}= 4 N - Acceleration of a box, a = ?

Applying the formula:

- a = F
_{net}/m - a = 4/2
- a = 2 m/s
^{2}

Therefore, a box accelerates forward at the rate of **2 m/s ^{2}**.

### Problem #2

Calculate the acceleration of a wooden cart weighing 12 kg when a net force of 6 N is applied to it.

**Solution**

Given data:

- Acceleration of a wooden cart, a = ?
- Mass of a wooden cart, m = 12 kg
- Net force applied to a wooden cart, F
_{net}= 6 N

Applying the formula:

- a = F
_{net}/m - a = 6/12
- a = 0.5 m/s
^{2}

Therefore, the acceleration of a wooden cart is **0.5 m/s ^{2}**.

### Problem #3

A wagon with a mass of 100 kg experiences a net force of 180 N. Determine the acceleration of the wagon.

**Solution**

Given data:

- Mass of a wagon, m = 100 kg
- Net force experienced by a wagon, F
_{net}= 180 N - Acceleration of a wagon, a = ?

Applying the formula:

- a = F
_{net}/m - a = 180/100
- a = 1.8 m/s
^{2}

Therefore, the acceleration of a wagon is **1.8 m/s ^{2}**.

### Problem #4

What is the acceleration of a horse cart weighing 400 kg when a net force of 800 N is applied to it?

**Solution**

Given data:

- Acceleration of a horse cart, a = ?
- Mass of a horse cart, m = 400 kg
- Net force applied to a horse cart, F
_{net}= 800 N

Applying the formula:

- a = F
_{net}/m - a = 800/400
- a = 2 m/s
^{2}

Therefore, the acceleration of a horse cart is **2 m/s ^{2}**.

## Related

## More topics

**Newton’s second law equation**- Newton’s law of universal gravitation formula
- Newton’s law of cooling equation
- Coulomb’s law equation
- Snell’s law equation
- Ohm’s law equation
- Hooke’s law equation

## External links

- https://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law
- https://study.com/learn/lesson/newton’s-second-law-of-motion-equation-examples.html
- https://www.grc.nasa.gov/www/k-12/rocket/newton2r.html
- https://www.studysmarter.us/explanations/math/mechanics-maths/newtons-second-law/
- https://www.omnicalculator.com/physics/newtons-second-law