AP Physics 1 · Topic 2.5

Topic 2.5: Newton's Second Law

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

Newton's second law states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system, and points in the same direction as that net force. Divide the net external force by the system's total mass to get it.

AP Physics: Unit 2 (topics 2.5 Newton's Second Law). AP Physics 1 Unit 2, Topic 2.5, covering learning objective 2.5.A, describe the conditions under which a system's velocity changes, and essential knowledge 2.5.A.1 through 2.5.A.3. The CED lists four suggested skills for this topic: 1.A, 2.A, 2.D, and 3.B. Unit 2 carries an 18 to 23 percent share of the multiple-choice section, matched only by Unit 3, with about 22 to 27 class periods suggested.

What the CED actually says

Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force. That is EK 2.5.A.2, and the relevant equation printed beside it is:

\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}

Notice the shape. Both the CED and the AP Physics 1 equation sheet solve for acceleration, not for force. That is not cosmetic: acceleration is the effect, and the fraction shows both dependencies at once. At fixed mass, doubling the net force doubles the acceleration; at fixed net force, doubling the mass halves it.

The direction clause is half the law and is the half most often dropped. The acceleration points along the net force, always, with no reference to which way the system happens to be moving. A car braking to a stop while traveling forward has a forward velocity and a backward acceleration, because the net force on it points backward.

Learning objective 2.5.A frames the topic as a question about motion rather than about forces: describe the conditions under which a system's velocity changes.

Unbalanced forces: the other half of Topic 2.4

EK 2.5.A.1 supplies the vocabulary. Unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero. Set that beside translational equilibrium in Topic 2.4 and the two topics are one dichotomy split across two pages.

Force configuration	Net force	Velocity of the system	CED objective
Balanced	Zero	Constant	2.4.A
Unbalanced	Nonzero	Changing	2.5.A

Two cautions travel with that table. A changing velocity is not the same as a changing speed: a net force perpendicular to the velocity turns it without altering the speed, which is what circular motion does in Topic 2.9. And balanced or unbalanced is decided axis by axis, so a system can be in equilibrium vertically while accelerating horizontally.

Neither topic says a force is needed to keep a system moving. What a net force changes is velocity, and only velocity.

Why the equation says center of mass

The subscripts in $\vec{a}_{\text{sys}}$ and $m_{\text{sys}}$ are the CED being precise about what accelerates. EK 2.5.A.3 completes the thought: the velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

External is the word that earns its keep. Wherever you draw the boundary, the forces inside it come in Newton's third law pairs, equal in magnitude and opposite in direction, acting on two members of the same system. Every such pair contributes zero to the sum, so internal forces, however violent, cannot accelerate the center of mass.

Two skaters at rest on frictionless ice push off each other and fly apart, yet the center of mass of the pair stays exactly where it was, because the push is internal and no external horizontal force acts.

Choosing the boundary well is therefore a real problem-solving move. Enclose two blocks and the connecting rope, and the tension disappears from the algebra; enclose one block and that same tension is external and has to be written down. The position of the center of mass comes from Topic 2.1 and is on the equation sheet:

\vec{x}_{\text{cm}} = \frac{\sum m_i \vec{x}_i}{\sum m_i}

The two forms on the equation sheet

Newton's second law appears twice in the mechanics column of the AP Physics 1 equation sheet. The Topic 2.5 form solves for acceleration. The other form solves for force and routes through momentum:

\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}

Read it left to right as a chain of claims. A net force is the rate at which a system's momentum changes. The middle step pulls the mass outside the change, which is legitimate while the system's mass stays constant. The last step is the familiar product.

Which form to reach for depends on what the problem hands you. Given forces and a mass, solve for the acceleration and continue with kinematics. Given a velocity change across a time interval, the momentum form is shorter, and it is the one the impulse-momentum theorem builds on in Unit 4.

They are the same law. If a question gives a force acting over a time and asks for a final speed, either route reaches the answer; the momentum form just skips a step.

One scalar equation per axis

The vector statement is executed one axis at a time:

\sum F_x = m a_x \qquad \sum F_y = m a_y

The procedure that fills those in, drawing the free-body diagram, fixing a positive direction, resolving each force into components, then summing with signs, is worked end to end in the how to find net force guide. Read that for the mechanics of the calculation, and check your arithmetic afterward with the net force calculator.

Two habits are worth naming. Choose the positive direction to match the direction you expect the acceleration to take and the signs mostly look after themselves; if the answer comes out negative, the system accelerates the other way and nothing is wrong. And on a ramp, tilt the axes to lie along and perpendicular to the surface, so the perpendicular equation reads $\sum F_\perp = 0$ and hands you the normal force. Inclined plane problems covers that setup in full.

Predicting factors of change

Suggested skill 2.D for this topic is predicting new values or factors of change of physical quantities using functional dependence between variables, and the second law is the cleanest place in the course to practice it. Because $a_{\text{sys}} = F_{\text{net}} / m_{\text{sys}}$ is a plain ratio, a whole class of questions can be answered without computing anything.

Change made	Net force	System mass	Acceleration
Push twice as hard	Doubles	Unchanged	Doubles
Cut the net force in half	Halves	Unchanged	Halves
Load the cart to triple its mass	Unchanged	Triples	One third
Double the force and double the mass	Doubles	Doubles	Unchanged
Double the force, quadruple the mass	Doubles	Quadruples	Halves

Two traps live in that table. Weight is proportional to mass, so a change that adds mass often changes some of the forces as well; check that the net force really held still before using a row. And it is the acceleration that scales, not the velocity. Doubling the net force on a moving system doubles the rate at which its velocity changes, which is a claim about the slope of the velocity graph, not about the speed at any given instant.

Four misreadings worth rehearsing

Acceleration is not velocity. A system can have zero velocity and a large acceleration, which is the situation at the instant a ball reverses at the top of a throw, or constant velocity and zero acceleration, which is Topic 2.4. Only the acceleration reports on the net force.
The net force is not a force. Nothing exerts it. It is the sum of the real forces, so it never gets an arrow on a free-body diagram, and neither does the product $m\vec{a}$ . See how to draw a free-body diagram.
Constant net force does not mean constant velocity. It means constant acceleration, so the velocity changes at a steady rate. That is exactly the condition the kinematic equations need on that axis; see Topic 1.2 for the definitions underneath.
More mass does not mean more acceleration. Mass sits in the denominator, so two systems driven by the same net force accelerate in inverse proportion to their masses. That is the CED's own essential question for Unit 2 about why a fully loaded dump truck is harder to stop than a small passenger car.

How Topic 2.5 is tested, and what to do next

Four suggested skills sit beside Topic 2.5 in the CED: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. The numbering maps onto the course's three science practices: 1.A is Creating Representations, 2.A and 2.D are Mathematical Routines, and 3.B is Scientific Questioning and Argumentation.

The CED's Unit 2 overview singles out 2.A and 2.D and connects them to the fourth free-response question, the Qualitative/Quantitative Translation. That question first asks for a claim with evidence and reasoning and no reference to equations, then for a derived equation or set of equations, then for the link between the two. Working the second law symbolically before any numbers go in is direct preparation.

At 18 to 23 percent of the multiple-choice section and roughly 22 to 27 class periods, Unit 2 shares the top weighting in the course with Unit 3. Next: gravitational force supplies the force this law is fed most often, and the Unit 2 overview maps the remaining topics.

Two blocks, one system, and the force between them

Two blocks sit in contact on a frictionless horizontal surface: a 2.0 kg block on the left and a 6.0 kg block on the right. A 32 N horizontal force pushes on the outer face of the 2.0 kg block, directed to the right. Take right as positive throughout. Find (a) the acceleration of the pair and (b) the contact force between the blocks.

(a) Choose the two-block system. Its mass is $m_{\text{sys}} = 2.0 + 6.0 = 8.0\ \mathrm{kg}$ . The contact force between the blocks is internal to this system, so it cancels against its Newton's third law partner and never enters the sum. The vertical forces balance, leaving the 32 N push as the only unbalanced external force.
Apply the second law to the system: $a_{\text{sys}} = \frac{F_{\text{net}}}{m_{\text{sys}}} = \frac{+32\ \mathrm{N}}{8.0\ \mathrm{kg}} = +4.0\ \mathrm{m/s^2}$ . Both blocks share this acceleration, because they stay in contact.
(b) Shrink the system to the 6.0 kg block alone. Redrawing the boundary makes the contact push external, and it is now the only horizontal force on that block: $F_c = m a = (6.0)(4.0) = 24\ \mathrm{N}$ , directed right.
Check against the other block. The 2.0 kg block feels $+32\ \mathrm{N}$ from the hand and $-24\ \mathrm{N}$ back from its neighbor, so $F_{\text{net}} = 32 - 24 = +8.0\ \mathrm{N}$ and $a = 8.0 / 2.0 = +4.0\ \mathrm{m/s^2}$ . Same acceleration, as it must be.
Now vary it. Apply the same 32 N to the outer face of the 6.0 kg block instead, pointing left, so $F_{\text{net}} = -32\ \mathrm{N}$ and $a_{\text{sys}} = -4.0\ \mathrm{m/s^2}$ . The contact force now has to accelerate only the 2.0 kg block: $(2.0)(4.0) = 8.0\ \mathrm{N}$ . Same masses, same applied force, same magnitude of acceleration, and a contact force three times smaller.

(a) $a_{\text{sys}} = +4.0\ \mathrm{m/s^2}$ , that is $4.0\ \mathrm{m/s^2}$ to the right. (b) The contact force is 24 N. The pair accelerates as a single 8.0 kg system driven by one external push, while the internal contact force only has to accelerate whichever block is not being pushed directly, which is why reversing the push drops it to 8.0 N.

Predicting a factor of change

A constant net force along a straight frictionless track gives a 1.6 kg cart an acceleration of $2.5\ \mathrm{m/s^2}$ . Take the direction the cart is pushed as positive. (a) Find the net force. (b) A 0.80 kg mass is added to the cart and the same net force is applied. Predict the new acceleration. (c) Starting from rest, how fast is the loaded cart moving 2.0 s after the force is applied?

(a) Rearrange the second law: $F_{\text{net}} = m_{\text{sys}} a_{\text{sys}} = (1.6\ \mathrm{kg})(2.5\ \mathrm{m/s^2}) = +4.0\ \mathrm{N}$ .
(b) Argue by factor of change first, which is what skill 2.D asks for. At fixed net force the acceleration is inversely proportional to mass. The mass goes from 1.6 kg to $1.6 + 0.80 = 2.4\ \mathrm{kg}$ , a factor of $2.4 / 1.6 = 1.5$ , so the acceleration falls by the same factor: $2.5 / 1.5 = 1.667\ \mathrm{m/s^2}$ .
Confirm by substitution: $a_{\text{sys}} = \frac{F_{\text{net}}}{m_{\text{sys}}} = \frac{4.0\ \mathrm{N}}{2.4\ \mathrm{kg}} = 1.667\ \mathrm{m/s^2}$ . Every input carries two significant figures, so report $+1.7\ \mathrm{m/s^2}$ .
(c) The net force is constant, so the acceleration is constant on this axis and the kinematic equations apply. From rest, $v = v_0 + a t$ .
Carry the unrounded acceleration into the arithmetic rather than the rounded one: $v = 0 + (1.667)(2.0) = 3.33\ \mathrm{m/s}$ , which is $+3.3\ \mathrm{m/s}$ to two significant figures. Rounding to $1.7\ \mathrm{m/s^2}$ first would have given 3.4 m/s, and the difference is entirely rounding.

(a) $F_{\text{net}} = +4.0\ \mathrm{N}$ . (b) $a_{\text{sys}} = +1.7\ \mathrm{m/s^2}$ , lower than 2.5 by the same factor of 1.5 that the mass rose by. (c) $v = +3.3\ \mathrm{m/s}$ after 2.0 s. All three answers follow from the single fact that at fixed net force, acceleration is inversely proportional to system mass.

Frequently asked questions

What does the AP Physics 1 equation sheet give for Newton's second law?

Two lines, both in the mechanics column. The first solves for acceleration: the acceleration of the system equals the sum of the forces divided by the mass of the system. The second routes through momentum: the net force equals the change in momentum divided by the time interval, which equals the mass times the change in velocity over that interval, which equals mass times acceleration.

Why is the second law written for a system's center of mass?

Because that is the point whose motion the law predicts for an extended or multi-part system. The CED states that the velocity of a system's center of mass changes only if a nonzero net external force is exerted on the system. Internal forces come in third law pairs that cancel inside the sum, so they never accelerate the center of mass, however energetic they are.

If the net force is constant, is the velocity constant?

No. A constant net force means a constant acceleration, so the velocity changes at a steady rate. That is exactly the condition the kinematic equations require on an axis. Velocity is constant only when the net force is zero, which is Newton's first law, Topic 2.4.

Does acceleration point the same way as velocity?

Only sometimes. Acceleration always points along the net force. When the two line up the system speeds up; when they oppose, it slows down; when the net force is perpendicular to the velocity, the direction turns while the speed holds. A braking car has forward velocity and backward acceleration.

Do internal forces change a system's acceleration?

No. A force exerted by one part of a system on another is matched by an equal and opposite force back on the first part, and both sit inside the sum, so the pair contributes nothing. Only external forces appear in the second law for that system. Redraw the boundary around one part and the same force becomes external, which is how contact forces and tensions are calculated.