C. Forces · Forces

Newton's Second Law: Fnet = ma

C2.2 C2.3 C3.3About 75 minutes

Learning goals

I can state Newton's second law and write it as Fnet = ma.
I can explain how acceleration depends on net force and on mass.
I can use free-body diagrams and Fnet = ma to find acceleration, net force, an unknown force or mass.
I can combine Newton's second law with the kinematics equations.

Words to know

Newton's second law
The acceleration of an object is in the direction of the net force, directly proportional to the net force and inversely proportional to the mass: a = Fnet/m, or Fnet = ma.

directly proportional
When one quantity doubles, the other doubles (a ∝ Fnet when m is constant).

inversely proportional
When one quantity doubles, the other halves (a ∝ 1/m when Fnet is constant).

acceleration (a)
The rate of change of velocity, in m/s². A vector with the same direction as the net force.

Push harder, or push less

Try pushing an empty grocery cart, then the same cart loaded with 20 kg bags of flour and potatoes. Push both with the same force: the empty cart zooms away, the full one barely speeds up. Now push the full cart twice as hard: it speeds up twice as fast.

The first law tells us what happens when the net force is zero. The second law tells us exactly what happens when it is not: how much the velocity changes, and in which direction.

Newton's second law

The acceleration of an object is directly proportional to the net force on it, inversely proportional to its mass, and in the direction of the net force.

a = Fnet ÷ m or Fnet = m a

Fnet in newtons (N), m in kilograms (kg), a in m/s². Rearranged: m = Fnet ÷ a.

Reading the equation

More net force, more acceleration. Double Fnet (same mass) and a doubles.
More mass, less acceleration. Double m (same Fnet) and a is halved.
Same direction. The acceleration always points the same way as the net force, not necessarily the way the object is moving. (A car braking moves forward but accelerates backward.)
Fnet = 0 gives a = 0. That is the first law: the first law is a special case of the second.

This also defines the newton: 1 N is the net force that gives a 1 kg mass an acceleration of 1 m/s², so 1 N = 1 kg·m/s².

The method

Draw an FBD and choose a positive direction (usually the direction of the acceleration, if you know it).
Find Fnet in that direction by adding the forces.
Use Fnet = ma. Solve for the unknown.
If needed, use the acceleration in a kinematics equation from the Kinematics unit: v₂ = v₁ + aΔt, Δd = v₁Δt + ½aΔt², v₂² = v₁² + 2aΔd.

Left: with mass fixed, doubling the net force doubles the acceleration (a straight line through the origin). Right: with the net force fixed, doubling the mass halves the acceleration (a curve that drops quickly, then levels off).

Worked example

Example 1: Finding acceleration

The net force on a 1500 kg car is 4500 N forward. Find its acceleration.

a = Fnet ÷ m = 4500 N ÷ 1500 kg
a = 3.0 m/s² forward (same direction as the net force).

3.0 m/s² forward

Worked example

Example 2: Finding force from motion

A 0.17 kg hockey puck is at rest. A slapshot gets it moving at 30 m/s in 0.010 s. Find the average net force on the puck.

First the acceleration (kinematics): a = (v₂ − v₁) ÷ Δt = (30 m/s − 0) ÷ 0.010 s = 3000 m/s².
Then the second law: Fnet = ma = (0.17 kg)(3000 m/s²) = 510 N in the direction of the shot.
That is about the weight of a 52 kg person, delivered in one hundredth of a second.

About 5.1 × 10² N, in the direction of the shot.

Worked example

Example 3: FBD, second law, then kinematics

A 25 kg crate is pulled across a level floor by a horizontal rope with a tension of 120 N. Kinetic friction is 70 N. (a) Find its acceleration. (b) If it starts from rest, how fast is it moving after 3.0 s, and how far has it moved?

FBD: Fg = 25 × 9.8 = 245 N down; FN = 245 N up; FT = 120 N right; Ff = 70 N left.
Right +: Fnet = 120 − 70 = 50 N.
(a) a = Fnet ÷ m = 50 ÷ 25 = 2.0 m/s² right.
(b) v₂ = v₁ + aΔt = 0 + (2.0)(3.0) = 6.0 m/s.
Δd = v₁Δt + ½aΔt² = 0 + ½(2.0)(3.0)² = 9.0 m.

(a) 2.0 m/s² right. (b) 6.0 m/s; 9.0 m.

The crate in Example 3. Vertically the forces balance. Horizontally, Fnet = 120 − 70 = 50 N right, so the crate accelerates to the right at 50 ÷ 25 = 2.0 m/s².

Worked example

Example 4: Finding an unknown force

A 60 kg cyclist (including the bike) slows from 8.0 m/s to 2.0 m/s in 4.0 s on level ground without pedalling. Air resistance and friction together are 30 N. She also brakes. Find the braking force.

a = (2.0 − 8.0) ÷ 4.0 = −1.5 m/s² (forward is +; the minus sign means backward).
Fnet = ma = 60 × (−1.5) = −90 N.
The backward forces are 30 N (resistance) plus the brake force Fb: Fnet = −30 − Fb = −90, so Fb = 60 N backward.

The brakes supply 60 N backward.

Just for fun

Newton described the link between force and changing motion in 1687. The first wheeled grocery cart was introduced in an Oklahoma supermarket in 1937, about 250 years later, finally giving everyone a way to test the second law every week.

Hands-on activity

Second law with a laundry basket

You need: an empty laundry basket (or a cardboard box), a smooth floor, a helper, some heavy items (books, large water jugs), and a phone with a stopwatch.

Mark a start line and a line 3 m away with tape.
Push the empty basket from rest, as steadily as you can with one hand (try to use the same, gentle push each time). Have your helper time it to the 3 m line. Repeat 3 times.
Add about 5 kg of books and repeat with the same gentle push. Then add another 5 kg.
From rest, Δd = ½aΔt², so a = 2Δd ÷ Δt². Calculate a for each trial and average each set.
Now keep the load fixed and push harder (two hands). What happens to a?
Write a conclusion: how did a change with mass and with force? List two reasons your results are not perfect (e.g. your push was not constant, friction).

Push gently and keep the area clear; never ride in the basket.

Your turn

A net force of 24 N acts on a 6.0 kg box. Find its acceleration.

What net force gives a 1200 kg car an acceleration of 2.5 m/s²?

A net force of 18 N gives an object an acceleration of 4.5 m/s². Find its mass.

The net force on an object is doubled and its mass is halved. What happens to its acceleration?

A 40 kg child on a toboggan is pulled with 90 N forward. Friction is 30 N. Find the acceleration. (The toboggan's mass is 5.0 kg.)

A car moving forward is braking. In which direction are its net force and acceleration?

More practice: warm-up

Write Newton's second law as an equation.

What are the units of each quantity in Fnet = ma?

Fnet = 10 N, m = 2.0 kg. Find a.

m = 3.0 kg, a = 2.0 m/s². Find Fnet.

If Fnet is zero, what is the acceleration?

More practice: on your own

A 0.45 kg soccer ball is kicked from rest to 18 m/s in 0.050 s. Find the average net force.

A 2.0 × 10³ kg truck accelerates from 10 m/s to 25 m/s in 6.0 s. Find the net force.

A 12 kg box is pushed with 80 N and accelerates at 4.0 m/s². Find the friction force.

A 50 kg sled is pulled from rest with a net force of 75 N. How far does it move in 4.0 s?

A 900 kg car slows from 20 m/s to rest over 40 m. Find the net braking force.

Challenge

A 1.0 × 10³ kg elevator starts moving up from rest and reaches 3.0 m/s in 2.0 s. Find the tension in the cable during this time.

The same net force gives object A an acceleration of 6.0 m/s² and object B an acceleration of 2.0 m/s². If A and B are tied together and the same force acts on both, what is the acceleration?

A 70 kg sprinter reaches 9.0 m/s in the first 2.0 s of a race on a track. Find the average horizontal force from the track on her shoes. What kind of force is this?

Check your understanding

Quiz

A 4.0 kg object has a net force of 20 N on it. Its acceleration is:

If the mass of an object is tripled and the net force stays the same, the acceleration:

The acceleration of an object always points in the direction of:

A 1500 kg car accelerates at 2.0 m/s². The net force on it is:

A 10 kg box is pulled with 50 N, and friction is 20 N. Its acceleration is:

Which equals 1 N?

A cart has a constant net force on it. Which graph best describes its velocity over time?

For parents and teachers

Big idea: Newton's second law, Fnet = ma (C2.2, C3.3), with the relationships a ∝ Fnet and a ∝ 1/m (C2.3). This is the central equation of the unit.

How to help: Insist on the routine: FBD, positive direction, Fnet, then F = ma. Check units at each step. A quick sense check: is the answer reasonable (a car's acceleration is a few m/s², not thousands)?

Common difficulties:

Using one force (e.g. the applied force) instead of the net force in F = ma.
Using weight in newtons where the mass in kilograms is needed.
Losing negative signs when an object slows down.
Assuming acceleration is in the direction of motion. It is in the direction of the net force.