Unit 2 · Topic 2.5

2.5 Newton’s Second Law

Newton's second law, $\vec{a}_{cm} = \frac{\sum\vec{F}}{m_{sys}}$ , connects forces to motion: a system's center of mass accelerates in the direction of the net external force. It's the main tool for every dynamics problem, from blocks on pulleys to forces that change with time.

Key terms

Newton's second law
net external force
acceleration
mass
connected objects

The second law

The acceleration of a system's center of mass is the net external force divided by the system's mass:

$\vec{a}_{cm} = \frac{\sum\vec{F}_{ext}}{m_{sys}}, \quad\text{or}\quad \sum\vec{F}_{ext} = m\vec{a}$

Three things follow. The acceleration points the same way as the net force (not necessarily the way the object is moving). Double the net force and the acceleration doubles. Double the mass and the acceleration halves.

Only external forces count. Forces between parts of the system cancel in pairs (2.3). And $m\vec{a}$ is not a force: never draw it on the FBD.

A routine for dynamics problems

Choose the system and draw its free-body diagram.
Choose axes, with one along the acceleration if you know its direction.
Write $\sum F_x = ma_x$ and $\sum F_y = ma_y$ with every force's component and sign.
Solve, then check units, signs and limiting cases (what if a mass is zero, or an angle is 0° or 90°?).

Connected objects

For objects joined by an ideal string, like an Atwood machine (two masses hanging over a pulley), write the second law for each object. They share the same size of acceleration, and the string's tension appears in both equations. Choose positive directions that follow the motion: if one mass goes down, the other goes up.

A shortcut: treat everything as one system moving along the string. The driving force is the weight of the heavier hanging mass minus the forces opposing it, and the mass is the total mass. For an Atwood machine with $m_2 > m_1$ :

$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$

Then go back to one object's equation to find the tension. The tension always lies between the two weights.

Net force and direction of motion

The net force sets the direction of the acceleration, not the direction of the velocity. A ball thrown upward is moving up, but the net force on it (gravity) points down, so it slows down. A car braking while moving forward has a backward net force.

So you can't read the forces from which way something moves. You read them from how its velocity is changing. A constant net force gives a constant acceleration, a straight-line velocity–time graph, and a parabolic position–time graph.

Weight and mass are different things. Mass, in kilograms, measures inertia and is the same everywhere. Weight, in newtons, is the gravitational force mg and depends on where you are.

When the force changes

In Physics C, forces may depend on time or position. Since $a = \frac{dv}{dt}$ , a time-dependent force gives $\frac{dv}{dt} = \frac{F(t)}{m}$ , which you integrate to get v(t), then integrate again for x(t). Velocity-dependent forces like drag lead to differential equations (2.9).

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Atwood machine
Masses of 3.0 kg and 5.0 kg hang on either side of an ideal pulley, connected by an ideal string. Find the acceleration and the tension. Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: The 5.0 kg mass goes down and the 3.0 kg mass goes up, each with acceleration a.
2. Step 2: 3.0 kg (up positive): $T - (3.0)(9.8) = 3.0a$ . 5.0 kg (down positive): $(5.0)(9.8) - T = 5.0a$ .
3. Step 3: Add the equations to eliminate T: $(5.0 - 3.0)(9.8) = 8.0a$ , so $a = 2.45$ m/s².
4. Step 4: Tension: $T = 3.0(9.8 + 2.45) \approx 36.8$ N. It's between the two weights (29.4 N and 49 N), as it must be.
Answer: a ≈ 2.45 m/s²; T ≈ 36.8 N.
Example 2Calculator allowed
A block on a table pulled by a hanging mass (classic trap)
A 4.0 kg block on a frictionless table is tied by a string over an ideal pulley at the table's edge to a 1.0 kg hanging mass. Find the acceleration and the tension. Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: Whole system along the string: the only unbalanced external force is the hanging mass's weight, 9.8 N. $a = \frac{9.8}{4.0 + 1.0} = 1.96$ m/s².
2. Step 2: Tension from the block alone (the only horizontal force on it): $T = (4.0)(1.96) \approx 7.8$ N.
3. Step 3: Check with the hanging mass: $9.8 - 7.84 = (1.0)(1.96)$ . ✓
4. Step 4: The trap is setting the tension equal to the hanging weight, 9.8 N. If it were, the hanging mass would have zero net force and couldn't accelerate.
Answer: a = 1.96 m/s²; T ≈ 7.8 N (less than 9.8 N).
Example 3
A force that grows with time
A 2.0 kg cart starts from rest on a frictionless track. A horizontal force $F(t) = 6t$ N (t in seconds) pushes it. Find its speed and how far it has moved at t = 2.0 s.
Show the solutionHide the solution
1. Step 1: $a(t) = \frac{F}{m} = 3t$ m/s². The acceleration isn't constant, so integrate.
2. Step 2: $v(t) = \int_0^t 3t\,dt = 1.5t^2$ . At t = 2.0 s, v = 6.0 m/s.
3. Step 3: $x(t) = \int_0^t 1.5t^2\,dt = 0.5t^3$ . At t = 2.0 s, x = 4.0 m.
Answer: 6.0 m/s and 4.0 m.

Common mistakes

Including $m\vec{a}$ as a force on the free-body diagram. It's the result of the forces, not one of them.
Setting the tension equal to a hanging mass's weight when the system accelerates.
Mixing sign conventions between connected objects. Choose positive directions that follow the motion for each object.
Using constant-acceleration equations when the force depends on time. Integrate a = F(t)/m.

On the exam

Free-response questions often ask you to derive an expression for acceleration or tension in terms of given letters (like $m_1$ , $m_2$ , θ and g). Start from $\sum F = ma$ for a clearly stated system and keep it symbolic until the end.
Expect "what happens if…" follow-ups: a mass doubles or an angle increases. Use your derived equation to justify the trend.

Connected topics

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Check yourself

4 questions on 2.5 Newton’s Second Law. Pick an answer to see if you got it, and why.

A 4.0 kg cart sits on a frictionless horizontal table. A light string tied to it runs over an ideal pulley at the table's edge to a hanging 1.0 kg block. The system is released from rest. Use g = 10 m/s².

Described situation

Question 1 of 4Calculator allowed

What is the acceleration of the cart?

Question 2 of 4Calculator allowed

What is the tension in the string?

Question 3 of 4Calculator allowed

A 2.0 kg object starts from rest on a frictionless surface. A horizontal force F = 6t acts on it, with F in newtons and t in seconds. What is the object's speed at t = 2.0 s?

Question 4 of 4Calculator allowed

A 3.0 kg object moves along the x-axis with position $x(t) = 2t^3$ , with x in meters and t in seconds. What is the net force on it at t = 1.0 s?

0 of 4 answered