Predict acceleration from net force and mass

Force changes velocity; it does not supply a velocity label

The cart begins at +2 m/s on a frictionless horizontal track. Set the applied horizontal force to zero. After one second it is at +2 m; after two seconds it is at +4 m. Its velocity is still +2 m/s. There is no horizontal acceleration, but there is motion.

This is the central distinction in Newton’s first law for an inertial frame: zero net force corresponds to constant velocity, including rest as one possible constant velocity. The initial condition determines which velocity persists. Everyday objects often slow when we stop pushing because other forces remain, not because maintaining velocity inherently consumes a forward force.

Add forces before dividing by mass

For the model’s constant mass, Newton’s second law is ΣF = ma. The symbol Σ means “sum”; add the force components acting on the selected object. The acceleration points with the net force. In one horizontal dimension, aₓ = ΣFₓ/m. The subscript x identifies the horizontal component; it is not an extra multiplying variable.

The OpenStax second-law treatment develops this connection. Here there is no horizontal drag or friction, so the applied horizontal force is also the horizontal net force. Vertically, the track and Earth forces cancel. The two component equations therefore give horizontal acceleration Fₓ/m and vertical acceleration zero.

For +4 N on 2 kg, acceleration is +2 m/s². Newtons divided by kilograms have units m/s², consistent with a rate of velocity change. This is not a velocity of 2 m/s. It means each second adds +2 m/s to velocity while the force and mass stay constant.

Follow the force through to a prediction

With initial position x₀ = 0 m, initial velocity v₀ = +2 m/s and constant acceleration a = +2 m/s², use the previous motion equations:

v(t) = v₀ + at and x(t) = x₀ + v₀t + ½at².

At t = 2 s, velocity is 2 + 2×2 = +6 m/s. Position is 0 + 2×2 + ½×2×4 = +8 m. The force model supplies a cause for acceleration; the initial conditions and duration then determine the motion. A force diagram alone does not supply initial velocity or elapsed time.

Your action: compare four captures with the same initial velocity and duration. At (force,mass) = (4 N,2 kg), acceleration is +2 m/s². At (8 N,2 kg), it doubles to +4 m/s². At (4 N,4 kg), it halves to +1 m/s². At (0 N,2 kg), it is zero. Predict the two-second velocities before inspecting the ledger: they are +6, +10, +4 and +2 m/s respectively.

Doubling force doubles acceleration when mass is fixed. Doubling mass halves acceleration when net force is fixed. These comparisons do not imply that final velocity doubles or halves, because the common initial velocity contributes as well. Keep the causal comparison and the arithmetic prediction separate.

A negative force can first slow, then reverse

Set force to −4 N and mass to 2 kg. Acceleration is −2 m/s², while initial velocity remains +2 m/s. The cart slows to zero velocity at one second and then moves negatively. At two seconds its velocity is −2 m/s and its position is zero.

The applied force does not turn off when the cart reaches zero velocity. It also does not flip direction when velocity reverses. That behavior is deliberately different from a friction model, whose direction depends on relative motion or impending slip. Treating a fixed negative force as “friction” would produce incorrect conclusions after the turn.

This case also shows why the force direction is not necessarily the current motion direction. Before one second the cart moves right while the net force points left. After one second both velocity and force point left. The same acceleration first reduces speed and then increases it in the opposite direction.

Respect the system and the frame

The equation ΣF = ma here uses the cart’s constant mass and an inertial ground frame. Its force sum excludes forces the cart exerts on Earth, the track and the agent. Equal-and-opposite third-law forces on other bodies do not cancel the cart’s net force. If the chosen system included multiple interacting bodies, internal and external forces would need to be reconsidered for that larger boundary.

The model assumes a horizontal frictionless track, constant applied force, fixed mass, no collision and no vertical acceleration. An accelerating observer, changing mass, an inclined force or speed-dependent drag would need a different analysis. The displayed values are calculated teaching predictions, not evidence that a real cart follows them. An experiment would need measured forces, mass, time and position, with uncertainties and competing explanations for discrepancies.

Transfer: A 5 kg cart begins at −1 m/s. Its horizontal forces are +9 N and −4 N, constant over three seconds; vertical forces balance. Find net force, acceleration, final velocity and displacement. Explain why adding the two force magnitudes would be wrong.

Check the net-force transfer

Net horizontal force is +9−4=+5 N, so acceleration is +1 m/s². Final velocity is −1+1×3=+2 m/s. Displacement is −1×3+½×1×9=+1.5 m. The forces oppose each other, so their signed components subtract. The object reverses during the interval, and displacement is not the total distance traveled.