Newton’s Second Law

Connect the external force sum to center-of-mass acceleration.

The net force changes the motion

Let’s push an empty shopping cart, then push the loaded cart with the same force. The loaded cart’s velocity changes more slowly. Push either cart harder and its velocity changes faster. Newton’s second law gives us the mathematical relationship behind that observation.

For a fixed-mass system in an inertial frame, ΣF⃗_external = M a⃗_CM. The acceleration of the system’s center of mass points in the direction of the net external force. Its magnitude is proportional to net force and inversely proportional to total mass. An unbalanced force configuration means the vector sum isn’t zero.

The direction of acceleration isn’t necessarily the direction of velocity. A cart can move right while the net force points left. It slows down at first, but may later reverse and speed up left if that force continues. Force predicts a change in velocity, not the current velocity itself.

A system boundary encloses a 3 kg rear cart and 2 kg front cart connected by a string. A 10 N external pull produces acceleration 2 m/s². The rear-cart equation gives tension 6 N.

Write the force sum, not just one force

For one axis, ΣFₓ = Ma_CM,x. Include all external components with signs. A 12 N pull right opposed by 4 N friction left gives net force +8 N, not +12 N. Balanced vertical forces don’t belong in the horizontal equation, but they still need their own force balance.

For a system of two connected carts, the forces they exert on each other are internal and cancel from the system sum. They still matter if you analyze one cart alone. Choosing the larger system often finds the common acceleration efficiently; choosing one part afterward can find the connecting tension.

Example 2.5.1 — A pair of carts

A 2 kg cart pulls a 3 kg cart along a frictionless horizontal track. An external 10 N force acts right on the 2 kg cart. The taut, ideal string keeps their acceleration magnitudes equal. For both carts together, total mass is 5 kg and net horizontal external force is 10 N. Thus a = 10/5 = 2.0 m/s² right.

For the 3 kg cart alone, the string supplies its only horizontal force: T = 3a = 6 N. For the 2 kg cart, 10 − T = 2a, and 10 − 6 = 4 N agrees with 2(2) = 4 N. The same T appears because the string is ideal; it doesn’t mean tension is external to the combined system.

Now add 2 N of opposing friction to each cart. The whole-system net force becomes 10 − 2 − 2 = 6 N, so acceleration is 6/5 = 1.2 m/s². For the rear 3 kg cart, T − 2 = 3(1.2), giving T = 5.6 N. Recalculate the force sum when the environment changes; don’t keep the previous tension by habit.

Use symbols to predict changes

If net force doubles with total mass fixed, acceleration doubles. If total mass triples with net force fixed, acceleration becomes one third as large. If both force and mass double, acceleration is unchanged. These claims hold only when the net force is actually held at the stated value.

Pulling a larger mass with the same hanging weight doesn’t automatically keep force on the cart constant: the hanging weight itself accelerates, so string tension can change. That is a reason to model both masses, or measure tension with a sensor rather than equating it to the hanging weight.

A graph can reveal the law

Use a force sensor and motion sensor on a low-friction cart. For several nearly constant pulls, measure net horizontal force and acceleration over the same time interval. Plot a versus F_net at fixed mass. The best-fit slope should be 1/M, with units kg⁻¹, and the intercept should be near zero if residual forces have been included.

Repeat with different total masses and a comparable measured net force, then plot a versus 1/M. Keep track slope, wheel behavior, and force-sensor calibration consistent. Repeated trials reduce random scatter; correcting a track tilt addresses a systematic effect. A meaningful procedure distinguishes those two kinds of improvement and explains what a slope tells you about the model.

A sliding incline block coupled to a hanging block

Assume a taut, negligible-mass, nonstretching rope and an ideal fixed pulley. Equal tension follows from these ideal assumptions, rather than from the system accelerating. The rope gives equal acceleration magnitudes along the two paths; the spatial acceleration vectors differ. While B slides downhill, friction acts uphill. Its uphill acceleration means it slows, not that it is already moving uphill; reconsider the friction regime when sliding stops.

Historical force/component construction diagrams. Gallery-6 combines gravity and projections; do not count projections as extra forces or present it as an AP exam FBD. Reference native dot-origin individual-force FBD instruction. Individual-force diagram conventions →

Notation and model qualifications in the retained source

Equal tension follows from the ideal rope/pulley, not from acceleration. Original wording →
The stray 40° attached to Block A belongs to Block B’s incline. Original wording →

Review the model and equations used here →

Example Problem 2.2.1

A 4 kg hanging block A is connected over an ideal fixed pulley to an 8 kg block B on a 40° incline. Block B is currently sliding down the incline. Use μₖ = 0.3 and g = 9.81 m/s². Find each block’s acceleration and its direction.

The negative acceleration means block B accelerates uphill while block A accelerates downward. If B is currently sliding downhill, it slows; negative acceleration alone does not mean it is already moving uphill.

Block B’s acceleration is 0.569 m/s² uphill; block A’s acceleration is 0.569 m/s² downward.

Therefore:

$\vec{T} = m_{A} {\vec{a}}_{A, y} + m_{A} \vec{g}$
The EOM for block B in the x direction is:
$Σ {\vec{F}}_{x} = m_{B} {\vec{a}}_{B, x}$
so:

${\vec{F}}_{g, B} \sin 40^{\circ} - {\vec{F}}_{f r} - \vec{T} = m_{B} {\vec{a}}_{x}$
and in the y- direction there is no acceleration so sum of forces in y-direction for block B is zero! Thus, we can set the normal force and the y component of the gravity force equal to each other.

${\vec{F}}_{N} = {\vec{F}}_{g, B} \cos 40^{\circ}$

For block B, there are forces acting in both the y and x-directions. In the x-direction we see friction, tension, and the x component of the weight. In the y-direction we have the normal force and y component of the weight .

Step 1: Set up your Free Body Diagram

Reference the image in this Gallery to view the FBD. Notice here that there are two FBDs constructed and two reference axes, one for block A, the other for block B. Each of the reference frames are oriented in the same direction as the object. Remember that we can set our reference axes to whatever orientation we want. It’ll cause less of a headache for us when we’re setting up our equations of motion.

Step 2: Break up the forces into x and y components

Based off the FBD, let’s approach the forces that are on each block. The force breakdown can be seen in this here Gallery. For block A, we see that forces are only in the y-direction which are tension and the weight of block A.

According to the reference axes for block A, the weight is acting in the negative y-direction so:

For block A, T − m_Ag = m_Aa, with upward positive. For the currently downhill-sliding block B, the shared signed acceleration is a = −0.56915 m/s².

Notice how the acceleration is negative. Due to our reference axes, we are assuming that a positive movement would be block B moving down the plane and pulling block A up.

Here are the following forces in equation form:

Block A: $40^{\circ}$

y-components:

${\vec{F}}_{T, A}$ ; ${\vec{F}}_{g, A} = m_{A} \vec{g}$

Block B:

x-components:

${\vec{F}}_{T, B}$ ; ${\vec{F}}_{g, B} \cdot \sin 40^{\circ}$ ; ${\vec{F}}_{f r} = μ_{k} \vec{N}$

y-components:

${\vec{F}}_{N}$ ; ${\vec{F}}_{g, B} \cos 40^{\circ}$

Step 3: Set up equations of motion

In order to solve for the acceleration of the system we have to approach each block individually. The EOM for block A is all in the y-direction so:

$Σ {\vec{F}}_{y} = m_{A} {\vec{a}}_{A, y}$

Now, we can rewrite the force balance as:

${\vec{F}}_{g, B} \sin 40^{\circ} - μ_{k} {\vec{F}}_{g, B} \cos 40^{\circ} - (m_{A} {\vec{a}}_{A} + m_{A} \vec{g}) = m_{B} {\vec{a}}_{B}$

Solving for a gives us:

${\vec{a}}_{B} = \frac{{\vec{F}}_{g, B} [\sin 40^{\circ} - μ_{k} \cos 40^{\circ}] - {\vec{F}}_{g, A}}{m_{B} + m_{A}}$
${\vec{F}}_{g, B} = 8 kg \cdot 9.81 m / s^{2} = 78.48 N$
${\vec{F}}_{g, A} = 4 kg \cdot 9.81 m / s^{2} = 39.24 N$
$μ_{k} = 0.3$

$m_{B}$ and $m_{A}$ are $8 kg$ and $4 kg$ respectively

With these given values, we can plug them into our equation to solve for acceleration.

${\vec{a}}_{B} = \overset{}{\frac{78.48 N (\sin 40^{\circ} - 0.3 \cos 40^{\circ}) - 39.24 N}{8 kg + 4 kg}}$

Notice how the Tension force is represented by just T for both blocks A and B. This is because we are concerned with how the system as a whole is accelerating so the tension force that is present in block A is the same as the tension force felt in block B. so we can rewrite the x-component of the force balance as:
${\vec{F}}_{g, B} \sin 40^{\circ} - μ_{k} {\vec{F}}_{N} - (m_{A} {\vec{a}}_{A, y} + m_{A} \vec{g}) = m_{B} {\vec{a}}_{B, x}$
and combine all the equations to get:
${\vec{F}}_{g, B} \sin 40^{\circ} - μ_{k} {\vec{F}}_{g, B} \cos 40^{\circ} - (m_{A} {\vec{a}}_{A, y} + m_{A} \vec{g}) = m_{B} {\vec{a}}_{B, x}$

Step 4: Solve for the acceleration of the system

The rope constrains the two blocks to have equal acceleration magnitudes along their respective paths. Their acceleration vectors point in different spatial directions.

Physics I by Jake Yap and Laurence Vincent. Complete historical context and correction history →

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Practice 1

A 4 kg cart has a 15 N rightward pull and 3 N leftward friction. Find acceleration and the velocity change after 2 s of constant force.

Full worked solution

Practice 2

For the two-cart example, derive a = F/(m₁ + m₂) and T = m₂F/(m₁ + m₂) on a frictionless track. Explain which system gives each equation.

Full worked solution

Practice 3

A best-fit graph of a versus F_net has slope 0.25 kg⁻¹. Infer the mass. Explain how an unaccounted constant friction force would affect the graph’s intercept.

Full worked solution

Keep exploring

The original Force Analysis and Common Forces preserve the connected-block problems. Analyze each with a stated system and external force sum.

Course scope and standards: the supplied AP Physics 1 CED, printed pages 49; Topic 2.5 and its exact learning objectives. The official College Board CED is the public source; the project audit records the supplied edition.

Supporting physics references: OpenStax: Newton’s second law and systems. These lessons, diagrams, examples, and practice questions were written for this book.

Newton’s Second Law

The net force changes the motion

Write the force sum, not just one force

Example 2.5.1 — A pair of carts

Use symbols to predict changes

A graph can reveal the law

A sliding incline block coupled to a hanging block

Example Problem 2.2.1

Step 1

Step 2

Example Problem 2.2.1 FBDs

Step 1

Example Problem 2.2.1 Components

Step 1

Step 2

Step 3

Check what you know

Let's see what stuck.

Give it a try

Practice 1

Practice 2

Practice 3

Keep exploring

Related source reading

Force

Velocity

Inertial frame

Acceleration

Vector component

Tension

Weight

Magnitude

Vector

Gravity

Free-body diagram

The net force changes the motion

Write the force sum, not just one force

Example 2.5.1 — A pair of carts

Use symbols to predict changes

A graph can reveal the law

A sliding incline block coupled to a hanging block

Example Problem 2.2.1

Step 1

Step 2

Example Problem 2.2.1 FBDs

Step 1

Example Problem 2.2.1 Components

Step 1

Step 2

Step 3

Check what you know

Give it a try

Practice 1

Practice 2

Practice 3

Keep exploring

Related source reading

Force

Velocity

Inertial frame

Acceleration

Vector component

Tension

Weight

Magnitude

Vector

Gravity

Free-body diagram

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