Aim for FivePhysics 1

Unit 2 · Topic 2.5

2.5 Newton’s Second Law

Newton's second law connects force to motion: the acceleration of a system's center of mass equals the net external force divided by its mass, a = ΣF/m, in the direction of the net force. Nearly every force problem in AP Physics 1 is solved by writing this law along each axis of a free-body diagram.

Key terms

  • Newton's second law
  • net force
  • unbalanced forces
  • mass
  • acceleration
  • inclined plane

The second law

When forces are unbalanced (ΣF ≠ 0), the system's center of mass accelerates: a_sys = ΣF/m_sys = F_net/m_sys. Doubling the net force doubles the acceleration; doubling the mass halves it.

The acceleration points in the direction of the net force, which isn't always the direction of motion. A car braking while moving east has a net force and acceleration pointing west.

Only external forces count. Forces between parts of the system cancel in pairs (2.3), so they can't change the velocity of the system's center of mass.

A step-by-step method

  • Choose the system: one object, or several objects that move together.
  • Draw a free-body diagram with every external force.
  • Choose axes with one axis along the acceleration (along the slope on an incline).
  • Write ΣF = ma for each axis. Forces along the acceleration are positive; forces against it are negative. If there's no acceleration along an axis, that sum is zero.
  • Solve in symbols first, then substitute numbers with units. Check that the answer makes sense in limiting cases.

Inclines

On a frictionless incline at angle θ, the only force along the slope is the component of gravity, mg sin θ. So ma = mg sin θ and a = g sin θ. The mass cancels, so every object slides down a frictionless incline with the same acceleration. Check the limits: θ = 0 gives a = 0 (flat ground) and θ = 90° gives a = g (free fall).

Connected objects

When objects are tied together by an ideal string, they share the same magnitude of acceleration. You can often find that acceleration fastest by treating everything as one system. Then the string tension is internal and drops out.

Example setup: a cart of mass m₁ on a frictionless table is tied over an ideal pulley to a hanging mass m₂. The only external force that drives the system is the hanging mass's weight, m₂g, and the total mass being accelerated is m₁ + m₂, so a = m₂g/(m₁ + m₂). To find the tension, go back to one object: for the cart, T = m₁a.

Two hanging masses on either side of a pulley (an Atwood machine) work the same way: a = (m₂ − m₁)g/(m₁ + m₂), with the heavier side going down.

Predicting changes and testing the law

AP questions often ask how acceleration changes when something is changed. Use your symbolic answer. In a = m₂g/(m₁ + m₂), m₂ appears in both the top and the bottom, so doubling m₂ doesn't double a.

In the lab, you can test the second law with graphs. Keep the mass fixed, vary the net force and plot a against F_net: you get a straight line through the origin whose slope is 1/m. Keep the force fixed and vary the mass: a against m is a curve, but a against 1/m is a straight line whose slope is F_net.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Cart pulled by a hanging mass (classic trap)

    A 3.0 kg cart on a frictionless table is connected by a light string over an ideal pulley to a 1.0 kg mass hanging over the edge. Use g = 9.8 m/s². Find the acceleration and the string tension.

    Show the solution
    1. Step 1: System = cart + hanging mass. External force along the motion: the hanging mass's weight, (1.0)(9.8) = 9.8 N. The tension is internal.
    2. Step 2: a = 9.8 N ÷ (3.0 + 1.0) kg = 2.45 m/s².
    3. Step 3: Cart alone: the only horizontal force is the tension, so T = m₁a = (3.0)(2.45) ≈ 7.35 N.
    4. Step 4: Check with the hanging mass: m₂g − T = m₂a gives 9.8 − 7.35 = 2.45 ✓. The trap is setting T = m₂g = 9.8 N. If the tension were that large, the hanging mass wouldn't accelerate.

    Answer: a ≈ 2.45 m/s²; T ≈ 7.35 N

  2. Example 2Calculator allowed

    Sliding down a frictionless ramp

    A block slides down a frictionless 20° incline. Derive an expression for its acceleration, then calculate it using g = 9.8 m/s².

    Show the solution
    1. Step 1: Forces: gravity (down) and the normal force (perpendicular to the incline). Choose x down the slope.
    2. Step 2: Along the slope: ΣF_x = mg sin θ = ma, so a = g sin θ.
    3. Step 3: Perpendicular: F_N − mg cos θ = 0, so F_N = mg cos θ (useful later for friction).
    4. Step 4: a = (9.8)(sin 20°) ≈ (9.8)(0.342) ≈ 3.35 m/s².

    Answer: a = g sin θ ≈ 3.4 m/s² down the incline, for any mass

  3. Example 3Calculator allowed

    Doubling the hanging mass

    In the cart-and-hanging-mass setup above (3.0 kg cart, 1.0 kg hanging mass), the hanging mass is replaced with a 2.0 kg mass. By what factor does the acceleration change? Use g = 9.8 m/s².

    Show the solution
    1. Step 1: Use a = m₂g/(m₁ + m₂).
    2. Step 2: New: a = (2.0)(9.8) ÷ (3.0 + 2.0) = 3.92 m/s².
    3. Step 3: Ratio: 3.92 ÷ 2.45 = 1.6. The driving force doubled, but the total mass also grew from 4.0 kg to 5.0 kg, so the acceleration increased by less than a factor of 2.

    Answer: The acceleration increases by a factor of 1.6 (to about 3.9 m/s²), not 2

Common mistakes

  • Including internal forces, like the tension between connected objects, when you treat the whole thing as one system.
  • Setting the tension equal to the hanging weight while the system accelerates.
  • Assuming the acceleration points the way the object moves. It points the way the net force points.
  • Plugging in numbers too early. A symbolic answer lets you check limits and answer "what if" questions.

On the exam

  • Mathematical Routines and Qualitative/Quantitative Translation questions often ask you to derive an expression for acceleration or tension "in terms of m, M and g." Start from ΣF = ma, and use only the given symbols.
  • Expect questions that ask how the acceleration changes if a mass or angle changes. Reason from your derived expression, and say which variable changes and how that affects the result.

Connected topics

Videos

Check yourself

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

Question 1 of 4Calculator allowed

A 2.0 kg block on a frictionless surface is pulled by a 10 N force to the right and a 4.0 N force to the left. What is its acceleration?

Question 2 of 4Calculator allowed

A block slides down a frictionless ramp tilted 30° above the horizontal. What is the size of its acceleration? Use g = 10 m/s².

A 10 kg box rests on a frictionless horizontal floor. A rope pulls on it with a 50 N force at 37° above the horizontal (sin 37° ≈ 0.60, cos 37° ≈ 0.80). Use g = 10 m/s².

Described situation

Question 3 of 4Calculator allowed

What is the box's acceleration?

Question 4 of 4Calculator allowed

What is the normal force from the floor on the box?

0 of 4 answered