Net Force Calculator (F = ma, Any Variable)

Enter any two of net force, mass, and acceleration and the calculator solves the third with Newton's second law, F = ma. A second mode adds two perpendicular force components to give the magnitude and direction of the net force. All inputs use SI units: N, kg, and m/s^2.

Mass m (kg)Acceleration a (m/s^2)

net force (F_net)

6 N

Steps

1.F_net = m a = 2 kg x 3 m/s^2 = 6 N
2.1 N = 1 kg m/s^2, so kg x m/s^2 gives newtons directly

AP Physics: Unit 2 (topics 2.5 Newton's Second Law). Newton's second law is Topic 2.5 in Unit 2 of AP Physics 1, which carries 18 to 23 percent of the multiple-choice section. AP Physics C: Mechanics covers the same topic in its Unit 2, weighted 20 to 25 percent.

What the calculator above does

The calculator above solves Newton's second law for whichever variable you leave blank. Enter any two of net force ( $F_{net}$ ), mass ( $m$ ), and acceleration ( $a$ ), and it returns the third using $F_{net} = ma$ . It also has a component mode: give it two perpendicular force components and it returns the magnitude and direction of their vector sum, which is the net force those components describe.

Everything runs in SI units: newtons for force, kilograms for mass, and meters per second squared for acceleration. If a problem hands you grams or kilometers per hour, convert before you type. The tool does the arithmetic, but you still have to identify the forces correctly, and that skill is covered step by step in how to find net force.

Newton's second law, solved for acceleration

The second law is often written solved for acceleration, which puts the cause and effect in the right order:

\vec{a} = \frac{\vec{F}_{net}}{m} = \frac{\sum \vec{F}}{m}

Reading it that way matters. Acceleration is the effect, and the net external force is the cause. The law applies to a system: add every external force acting on the system, divide by the system's total mass, and you get the acceleration of the system's center of mass. Forces that objects inside the system exert on each other cancel in third-law pairs, so they never show up in the sum.

$F_{net} = ma$ is the same equation rearranged, and the calculator handles every arrangement. Just make sure the force you enter is the net force on the system, not one applied force acting alone.

Solving for each variable

Each version of the equation answers a different question:

Force: enter mass and acceleration. A 1500 kg car accelerating at $2.0 \text{ m/s}^2$ needs $F_{net} = 1500 \times 2.0 = 3000$ N.
Mass: enter net force and acceleration, and the tool computes $m = F_{net}/a$ . This is how you find the mass of a cart from a force probe and a motion sensor in a lab.
Acceleration: enter net force and mass to get $a = F_{net}/m$ , then carry that acceleration into the kinematics calculator if the problem asks for velocity or position.

Pick a positive direction before you enter anything. A force pointing the opposite way goes in with a minus sign, and a negative acceleration in the output means the acceleration points opposite your chosen direction.

Adding two force components

Component mode handles the case where forces act along different axes. Perpendicular components add with the Pythagorean theorem:

F_{net} = \sqrt{F_x^2 + F_y^2}

and the direction comes from $\theta = \tan^{-1}(F_y / F_x)$ , measured from the x axis.

Where do the components come from? A free-body diagram. Draw every force acting on the object, resolve any angled force into x and y parts, sum each axis separately, and feed the two axis totals into the calculator. If your diagrams feel shaky, work through how to draw a free-body diagram first, then practice in the free-body diagram builder. On the AP exam, a clean diagram is usually worth points on its own before you ever reach the algebra.

Mistakes the calculator cannot catch

The tool trusts your inputs, so these errors pass straight through:

Entering one force instead of the net force. If you push a crate with 60 N and friction pulls back with 24 N, the second law uses 36 N, not 60 N.
Confusing weight with mass. Mass goes in as kilograms. Weight is a force: $w = mg$ . A problem that gives a weight of 490 N is telling you the mass is $490 / 9.8 = 50$ kg.
Dropping a sign. Two opposing forces must carry opposite signs before you add them.
Skipping the perpendicular forces. On a flat surface the normal force and gravity cancel, but on an incline they do not. See how to find normal force and the friction calculator for those setups, and how to find tension when ropes or pulleys pull on the object.

Where this sits in the AP course

Newton's second law is Topic 2.5 in Unit 2, Force and Translational Dynamics, which carries 18 to 23 percent of the AP Physics 1 multiple-choice section, tied with Unit 3 for the largest share of any unit. In AP Physics C: Mechanics, the same unit is weighted even higher at 20 to 25 percent.

A four-function, scientific, or graphing calculator is allowed on both sections of the exam, so the arithmetic you practice here transfers directly. What the exam really tests is setup: choosing the system, drawing the diagram, and writing the sum of forces before any numbers appear. Review the whole unit at Unit 2: Force and Translational Dynamics and keep the AP Physics 1 formula sheet open while you work.

Solve for acceleration with two opposing forces

You push a 12.0 kg crate across a floor with a horizontal force of 60.0 N. Friction opposes the motion with a force of 24.0 N. What is the crate's acceleration?

Take the direction of the push as positive. Sum the horizontal forces: $F_{net} = 60.0 \text{ N} - 24.0 \text{ N} = 36.0 \text{ N}$ .
Apply the second law solved for acceleration: $a = F_{net}/m = 36.0 \text{ N} / 12.0 \text{ kg}$ .
Divide: $a = 3.00 \text{ m/s}^2$ . The sign is positive, so the acceleration points in the direction of the push.

$a = 3.00 \text{ m/s}^2$ in the direction of the 60.0 N push. Entering 36.0 N and 12.0 kg in the calculator above returns the same result.

Solve for mass from force and acceleration

A net force of 45.0 N gives a loaded sled an acceleration of 1.80 m/s^2. What is the mass of the sled?

Rearrange $F_{net} = ma$ to isolate mass: $m = F_{net}/a$ .
Substitute: $m = 45.0 \text{ N} / 1.80 \text{ m/s}^2 = 25.0 \text{ kg}$ .
Check that the answer is reasonable: the sled's weight would be $w = mg = 25.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 245 \text{ N}$ , so a 45.0 N net force producing a modest acceleration makes sense.

$m = 25.0 \text{ kg}$ .

Sum two components, then find acceleration

Two forces act on a 2.00 kg cart on a frictionless table: 6.00 N east and 8.00 N north. Find the magnitude and direction of the net force, then the cart's acceleration.

The forces are perpendicular, so use the Pythagorean theorem: $F_{net} = \sqrt{(6.00)^2 + (8.00)^2} = \sqrt{36.0 + 64.0} = \sqrt{100.0} = 10.0 \text{ N}$ .
Find the direction: $\theta = \tan^{-1}(8.00/6.00) = \tan^{-1}(1.33) = 53.1^\circ$ north of east.
Apply the second law: $a = F_{net}/m = 10.0 \text{ N} / 2.00 \text{ kg} = 5.00 \text{ m/s}^2$ .
Acceleration points in the same direction as the net force: $53.1^\circ$ north of east.

$F_{net} = 10.0$ N at $53.1^\circ$ north of east, giving $a = 5.00 \text{ m/s}^2$ in that same direction.

Frequently asked questions

Should I write Newton's second law as F = ma or a = Fnet/m?

Both are the same relationship rearranged, so use whichever isolates the variable you want. Writing it solved for acceleration, a = Fnet/m = (sum of F)/m, keeps the physics in order: the net external force on a system causes the acceleration of the system's center of mass.

What units does the net force calculator use?

SI units throughout: newtons (N) for force, kilograms (kg) for mass, and meters per second squared for acceleration. One newton equals one kilogram meter per second squared, so the units close on their own. Convert grams to kilograms before entering a mass.

What is the difference between net force and applied force?

An applied force is one individual push or pull. The net force is the vector sum of every force acting on the object: applied forces, friction, gravity, normal force, tension, all of them. Newton's second law only works with the net force, which is why finding the net force from a free-body diagram comes first.

Can the net force be zero while the object is moving?

Yes. Zero net force means zero acceleration, not zero velocity. An object moving at constant velocity has balanced forces acting on it. That case belongs to Newton's first law, Topic 2.4 in the AP Physics 1 course, and it trips up a lot of students on the exam.

Is weight the same thing as mass?

No. Mass is the quantity in kilograms that goes into F = ma. Weight is the gravitational force on that mass, w = mg, using g = 9.8 m/s^2 near Earth's surface. A 10.0 kg object weighs 98 N. If a problem gives you weight in newtons, divide by 9.8 to get the mass before using the calculator.