Visighta

Applications of Translational Dynamics

6.4 Nonconstant Net Force and Acceleration as a State-Dependent Quantity

The Acceleration That Changes Itself

A rocket lifts off the pad. Its engines produce a constant thrust --- the same force, second after second. And yet the rocket accelerates faster and faster as it climbs. Why?

Because the rocket is burning fuel. As propellant is ejected, the rocket's mass decreases. The thrust stays the same, but $a = F/m$, and $m$ is shrinking. The acceleration increases over time even though the applied force does not change. The acceleration at any moment depends on the current state of the system --- in this case, how much fuel remains.

This is not an exotic special case. It is the generic situation in mechanics. A spring pushes harder the more it is compressed. Gravity weakens as you move away from Earth. Drag increases as you speed up. In each case, the force --- and therefore the acceleration --- depends on where the object is, how fast it is moving, or how the system is configured right now. The acceleration is not a fixed number assigned at the start of the problem. It is a quantity that is produced, moment by moment, by the current conditions.

Sections 6.1 through 6.3 dealt with problems where the net force was either constant or had a simple dependence on velocity. This section confronts the general case head-on: what happens when you cannot assign a single value to "the acceleration"?

Before you read on: A ball is pressed against a horizontal spring, compressing it. The spring is released and the ball accelerates along a frictionless surface until it loses contact with the spring.

During the time the ball is in contact with the spring, does the ball's acceleration increase, decrease, or stay constant?

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. The student selects "increases," "decreases," or "stays constant." Their choice is locked in and revisited after the exploration.]

Exploration: A Mass on a Spring

The best way to see state-dependent acceleration in action is to watch it unfold. The following interactive shows a mass attached to a horizontal spring, free to oscillate on a frictionless surface.

[Interactive: Spring Oscillator Lab. A mass $m$ is attached to a spring with adjustable stiffness $k$. The mass oscillates horizontally on a frictionless surface. Three time-series graphs are displayed simultaneously: position $x(t)$, velocity $v(t)$, and acceleration $a(t)$. Below the graphs, a force arrow and an acceleration arrow are drawn on the mass, updating in real time.

The student can adjust: - Spring constant $k$ (slider, 1 to 20 N/m) - Mass $m$ (slider, 0.5 to 5 kg) - Initial displacement $x_0$ (slider, $-0.5$ to $0.5$ m)

Controls: Play/Pause, Step (advance one small time increment), Reset.

Guided prompts appear as the student explores:]

Step 1: Set $k = 5$ N/m, $m = 1$ kg, and $x_0 = 0.3$ m. Press Play and watch the mass oscillate. Look at the three graphs.

Where is the acceleration largest in magnitude? Where is the velocity largest in magnitude? Are they at the same place?

Step 2: Pause the simulation when the mass passes through the equilibrium position ($x = 0$). What are the force, acceleration, and velocity at this instant?

Step 3: Pause the simulation at the point of maximum displacement (either side). What are the force, acceleration, and velocity now?

Step 4: Increase the spring constant to $k = 15$ N/m and replay. What changes about the acceleration? What stays the same about the relationship between $a$ and $x$?

If you worked through those explorations carefully, you discovered a pattern:

  • The acceleration is largest at the endpoints, where the displacement is greatest and the spring force is strongest.
  • The velocity is largest at the center ($x = 0$), where the acceleration is zero.
  • The acceleration and velocity reach their maximum magnitudes at different locations. This is one of the most counterintuitive features of oscillatory motion.
  • Increasing $k$ makes the acceleration larger at every point, but the relationship $a = -(k/m)x$ still holds at each instant.

The acceleration is not "large" or "small" as a property of the motion. It is large here and small there, determined at each position by the spring force at that position.

The Concept: Acceleration Determined by Current State

Here is the central idea, stated plainly.

When the net force on an object depends on its position, velocity, or configuration, the acceleration is not a single number that characterizes the entire motion. It is a quantity that changes from moment to moment, determined by the current state of the system through Newton's second law:

$$a = \frac{F_{\text{net}}(x, v, \text{configuration})}{m}$$

At each instant, you look at where the object is and how it is moving, compute the net force from the force laws, and divide by the mass. That gives you the acceleration right now. A moment later, the object has moved, the state has changed, the force is different, and the acceleration is different too.

This is the physical content of the differential equation $F(x, v) = ma$ that Section 4.1 formulated mathematically. There, you learned that this equation generates motion step by step from initial conditions. Here, you see it from the physics side: the force changes because the situation changes, and the acceleration follows the force.

Three important consequences follow:

1. There is no single "the acceleration." Asking "what is the acceleration?" for a mass on a spring is like asking "what is the temperature?" without specifying a location on Earth. The acceleration depends on where the object is in its motion. You can talk about the acceleration at a particular position or at a particular instant, but not "the acceleration" of the whole trajectory.

2. Constant-acceleration formulas do not apply. The kinematic equations $v = v_0 + at$ and $x = x_0 + v_0 t + \frac{1}{2}at^2$ were derived under the assumption that $a$ is constant. When $a$ varies with position or velocity, these formulas give wrong answers. Using them is like using a flat-Earth map for intercontinental navigation --- fine over short distances, but increasingly wrong as the journey grows.

3. The acceleration is produced locally. The spring does not "know" what the acceleration was a second ago or what it will be a second from now. It pushes based on its current compression. Gravity pulls based on the current distance. Each force responds to the present state and nothing else. The acceleration at each moment is a local, instantaneous consequence of current conditions.

Connection to the Differential Equation Framework

In Section 4.1, you learned to translate "the force depends on the state" into a differential equation. For a spring:

$$m\frac{d^2x}{dt^2} = -kx$$

For a falling object with drag:

$$m\frac{dv}{dt} = mg - bv$$

At the time, the emphasis was on the mathematical structure --- how the unknown appears inside its own equation, why you need initial conditions, what it means to solve an initial-value problem.

Now you are seeing the same idea from the other direction. The differential equation is not an abstract formalism. It is a literal description of what happens physically: the force depends on the state, so the acceleration depends on the state, so you must recompute it at every moment. The equation $F(x, v) = ma$ is just Newton's second law applied honestly, without the simplifying assumption that $a$ is constant.

The mathematical framework of Chapter 4 and the physical intuition of this section are two views of the same thing.

Return to the Prediction

The ball is pressed against a spring and released. During contact, does the acceleration increase, decrease, or stay constant?

The answer: the acceleration decreases.

At the moment of release, the spring is maximally compressed. The spring force is at its largest, so the acceleration is at its largest. As the ball moves forward, the spring extends back toward its natural length. The compression decreases, the spring force decreases, and the acceleration decreases. At the instant the ball separates from the spring (when the spring reaches its natural length), the spring force is zero and the acceleration is zero.

The ball's velocity, meanwhile, increases throughout this process --- the ball speeds up even as the acceleration decreases. This is not a contradiction. The acceleration is positive the entire time (the spring pushes forward), so the velocity keeps growing. But the rate at which the velocity grows diminishes because the force is weakening.

After the ball loses contact, it moves at constant velocity on the frictionless surface --- no force, no acceleration, constant speed.

If you predicted "increases," you may have been thinking that a faster ball should have a larger acceleration. But that reverses the causality. The acceleration determines how the velocity changes, not the other way around. A decreasing acceleration can still produce an increasing velocity, as long as the acceleration remains positive.

Worked Example: Acceleration at Several Positions

A 2 kg block slides along a frictionless surface and encounters a spring ($k = 200$ N/m) at $x = 0$. The block compresses the spring to a maximum compression of $x = -0.1$ m before bouncing back. Find the acceleration at $x = 0$, $x = -0.03$ m, $x = -0.06$ m, and $x = -0.1$ m.

At each position, the spring force is $F = -kx$ (taking leftward/compression as negative $x$):

At $x = 0$: $F = -200(0) = 0$ N, so $a = 0/2 = 0$ m/s$^2$.

At $x = -0.03$ m: $F = -200(-0.03) = 6$ N (to the right), so $a = 6/2 = 3$ m/s$^2$.

At $x = -0.06$ m: $F = -200(-0.06) = 12$ N, so $a = 12/2 = 6$ m/s$^2$.

At $x = -0.1$ m: $F = -200(-0.1) = 20$ N, so $a = 20/2 = 10$ m/s$^2$.

Notice the pattern: the acceleration is proportional to the displacement. The relationship is $a = -(k/m)x = -100x$ at every point. At maximum compression, the acceleration is at its maximum. At equilibrium, the acceleration is zero. This is not a coincidence --- it is a direct consequence of Hooke's law plus Newton's second law.

Also notice: the velocity is zero at $x = -0.1$ m (the turnaround point) and maximum at $x = 0$ (where the block first contacts the spring). Maximum acceleration and maximum velocity occur at opposite ends of the motion.

Faded Example: Gravitational Force Near Earth's Surface

A small asteroid of mass $m$ falls toward Earth from a height $h$ above the surface, where $h$ is large enough that the variation of gravity matters. The gravitational acceleration at distance $r$ from Earth's center is:

$$a(r) = -\frac{GM}{r^2}$$

where $G$ is the gravitational constant and $M$ is Earth's mass. Take the positive direction as radially outward.

Step 1: Find the acceleration at the surface ($r = R_E$, Earth's radius).

Check your answer $$a(R_E) = -\frac{GM}{R_E^2} = -g \approx -9.8 \text{ m/s}^2$$ This is just the familiar surface gravity, directed inward.

Step 2: Find the acceleration at $r = 2R_E$ (one Earth radius above the surface).

Check your answer $$a(2R_E) = -\frac{GM}{(2R_E)^2} = -\frac{GM}{4R_E^2} = -\frac{g}{4} \approx -2.45 \text{ m/s}^2$$ At twice the distance from Earth's center, the gravitational acceleration is one-quarter of the surface value. Gravity weakens rapidly with distance.

Step 3: Why can't you use $x = x_0 + v_0 t + \frac{1}{2}at^2$ to find the time for the asteroid to fall from $r = 2R_E$ to the surface?

Check your answer Because $a$ is not constant during the fall. At the start ($r = 2R_E$), the acceleration is $g/4$. At the end ($r = R_E$), the acceleration is $g$. Using any single value of $a$ --- whether the starting value, the ending value, or the average --- in the constant-acceleration formula would give an incorrect result. The correct approach is to solve the differential equation $d^2r/dt^2 = -GM/r^2$ with appropriate initial conditions, which is exactly what Section 4.1 prepared you to do.

Sketching $a(x)$: Seeing the Acceleration Profile

One of the most useful skills for understanding nonconstant acceleration is drawing $a(x)$ --- a graph of how the acceleration varies with position. This is not the same as $a(t)$. Instead of asking "what is the acceleration at each moment in time?", you are asking "what is the acceleration at each location in space?"

For forces that depend on position, $a(x)$ has a simple relationship to $F(x)$:

$$a(x) = \frac{F(x)}{m}$$

The acceleration profile is just the force profile scaled by $1/m$. If you can sketch $F(x)$, you can sketch $a(x)$.

Spring force: $F(x) = -kx$. This is a straight line through the origin with negative slope. The acceleration profile $a(x) = -(k/m)x$ is also a straight line through the origin. At positive $x$, the acceleration is negative (pushes back toward the center). At negative $x$, the acceleration is positive (pushes back toward the center). The magnitude of $a$ grows linearly with displacement.

Gravity near Earth (radial): $F(r) = -GMm/r^2$, so $a(r) = -GM/r^2$. This is a curve that is steeply negative near $r = R_E$ and flattens out as $r$ increases. The acceleration weakens with distance, falling off as $1/r^2$.

Constant force (for comparison): $F = mg$, so $a = g$ everywhere. The acceleration profile is a horizontal line. This is the only case where constant-acceleration kinematics applies exactly.

Pause and think: Sketch $a(x)$ for a spring and for constant gravity on the same axes. Where do they agree? Where do they differ? What does the shape of each curve tell you about the motion?

[Interactive: Acceleration Profile Plotter. The student selects a force law from a dropdown: Spring ($F = -kx$), Gravity ($F = -GMm/r^2$), Constant ($F = mg$), or Linear-plus-constant ($F = mg - kx$). Adjustable parameters appear for each case. The display shows $F(x)$ and $a(x) = F(x)/m$ on the same plot.

Guided prompt: "Compare the spring and constant-force profiles. For what range of $x$ is the spring's acceleration close to a constant value? Where does the approximation break down?"]

Why Constant-Acceleration Formulas Fail

This is worth stating directly, because the temptation to use $v^2 = v_0^2 + 2a\Delta x$ or $x = x_0 + v_0t + \frac{1}{2}at^2$ is strong.

Those formulas were derived in Chapter 2 under a specific assumption: that $a$ is the same number at every point in the motion. When that assumption holds, the formulas are exact. When it does not hold, they are wrong.

Consider a mass on a spring. At $x = 0.1$ m, the acceleration might be $-10$ m/s$^2$. At $x = 0.05$ m, it is $-5$ m/s$^2$. At $x = 0$, it is $0$. If you plug $a = -10$ m/s$^2$ into a constant-acceleration formula, you are assuming that the acceleration stays at $-10$ m/s$^2$ for the entire motion. But it does not --- it changes at every position. The formula will overestimate how quickly the object decelerates and predict a trajectory that is too "sharp."

The correct tools for variable acceleration are:

  • Differential equations (Chapter 4): solve $F(x, v) = ma$ directly.
  • Energy methods (Chapter 7, coming soon): use conservation of energy to relate speed and position without solving the differential equation.
  • Numerical methods: step through the motion in small increments, recomputing $a$ at each step.

The constant-acceleration formulas are a special case of these general methods --- the special case where $F$ does not depend on $x$ or $v$.

Practice

Layer 1: Concrete

Problem 1. A 0.5 kg object moves along a line under the influence of a position-dependent force $F(x) = -8x$ (in newtons, with $x$ in meters).

(a) Find the acceleration at $x = 0$, $x = 0.2$ m, $x = 0.5$ m, and $x = -0.3$ m.

(b) At which of these positions is the object speeding up, and at which is it slowing down? (You need additional information to answer this. What information do you need?)

Check your answer **(a)** Using $a = F/m = -8x / 0.5 = -16x$: - At $x = 0$: $a = 0$ m/s$^2$. - At $x = 0.2$ m: $a = -16(0.2) = -3.2$ m/s$^2$. - At $x = 0.5$ m: $a = -16(0.5) = -8.0$ m/s$^2$. - At $x = -0.3$ m: $a = -16(-0.3) = +4.8$ m/s$^2$. **(b)** You cannot determine whether the object is speeding up or slowing down from the acceleration alone --- you also need to know the direction of the velocity. The object speeds up when the acceleration and velocity point in the same direction, and slows down when they point in opposite directions. For example, at $x = 0.2$ m with $a = -3.2$ m/s$^2$: if the object is moving in the positive $x$-direction, it is slowing down (acceleration opposes velocity). If it is moving in the negative $x$-direction, it is speeding up (acceleration and velocity are both negative). This is a reminder that acceleration and "speeding up" are not the same thing.

Problem 2. A 1500 kg car experiences a braking force that depends on its speed: $F(v) = -600v$ (in newtons, with $v$ in m/s).

(a) What is the acceleration when $v = 20$ m/s? When $v = 5$ m/s?

(b) Does the car decelerate uniformly? Explain.

Check your answer **(a)** Using $a = F/m = -600v / 1500 = -0.4v$: - At $v = 20$ m/s: $a = -0.4(20) = -8$ m/s$^2$. - At $v = 5$ m/s: $a = -0.4(5) = -2$ m/s$^2$. **(b)** No. The deceleration is larger when the car is moving fast and smaller when the car is moving slowly. The car brakes hard at first and then progressively more gently. This is characteristic of velocity-dependent resistive forces --- the braking diminishes as the speed drops, and the car takes longer to come to rest than it would under constant deceleration. (In fact, the car asymptotically approaches rest, never quite reaching $v = 0$ in finite time, as you saw in Section 4.1.)

Layer 2: Pattern

Problem 3. For each force profile below, sketch $a(x)$ on the same set of axes. Use $m = 1$ kg for simplicity.

  • (a) $F(x) = -kx$ with $k = 4$ N/m (spring)
  • (b) $F = -mg$ with $g = 10$ m/s$^2$ (constant gravity, downward)
  • (c) $F(x) = -kx + mg$ with $k = 4$ N/m and $mg = 10$ N (spring with gravity --- a vertical spring supporting a weight)
Check your answer **(a)** $a(x) = -4x$. A straight line through the origin, sloping downward. At $x = 1$, $a = -4$; at $x = -1$, $a = 4$. **(b)** $a = -10$ m/s$^2$ everywhere. A horizontal line at $a = -10$, independent of $x$. **(c)** $a(x) = -4x + 10$. A straight line with the same slope as (a), but shifted upward by 10. The acceleration is zero at $x = 10/4 = 2.5$ m --- this is the new equilibrium position. Below this point, the acceleration is positive (spring and gravity together push the mass upward); above it, the spring is stretched enough that it overcomes gravity and pulls the mass back down. Notice that (c) is just (a) shifted. The spring-plus-gravity system oscillates around a new equilibrium, not around $x = 0$. The *shape* of $a(x)$ --- a straight line with negative slope --- is the same in both cases. What changes is where the line crosses zero.

Problem 4. A particle moves along the $x$-axis under a force that produces the acceleration profile shown below (described verbally, since this is text):

  • For $x < 0$: $a = +2$ m/s$^2$ (constant, positive).
  • For $0 \leq x \leq 1$ m: $a$ decreases linearly from $+2$ to $-2$ m/s$^2$.
  • For $x > 1$ m: $a = -2$ m/s$^2$ (constant, negative).

(a) At what position is $a = 0$?

(b) Describe qualitatively what happens to a particle released from rest at $x = -1$ m.

Check your answer **(a)** The acceleration decreases linearly from $+2$ at $x = 0$ to $-2$ at $x = 1$ m. It passes through zero at $x = 0.5$ m. **(b)** At $x = -1$ m, the acceleration is $+2$ m/s$^2$. The particle accelerates to the right. It continues to accelerate as long as $a > 0$, which holds for $x < 0.5$ m. At $x = 0.5$ m, the acceleration is zero, but the velocity is at its maximum (the particle has been speeding up the entire time). Past $x = 0.5$ m, the acceleration is negative --- the particle decelerates. It slows down, eventually stops, and is pushed back toward $x = 0.5$ m. The particle oscillates around the equilibrium at $x = 0.5$ m. This is the same qualitative behavior as a mass on a spring, even though the force profile looks different in the outer regions. The key feature is the negative slope of $a(x)$ near equilibrium --- this is what produces oscillation.

Layer 3: Structure

Problem 5. Why can't you use constant-acceleration formulas when the net force depends on position?

Check your answer The constant-acceleration formulas ($v = v_0 + at$, $x = x_0 + v_0t + \frac{1}{2}at^2$, $v^2 = v_0^2 + 2a\Delta x$) are derived by integrating $a = \text{const}$ over time. The derivation *requires* that $a$ does not change during the motion. When the net force depends on position, $a$ changes as the object moves. As the object travels from one point to another, the force at the new point is different from the force at the starting point, so the acceleration is different. Using a single value of $a$ in the kinematic formulas amounts to pretending the force does not change --- which contradicts the problem setup. More precisely: the derivation of $x = x_0 + v_0 t + \frac{1}{2}at^2$ assumes $\int_0^t a\,dt' = at$, which is only true when $a$ is constant. When $a = a(x(t'))$, this integral depends on the trajectory $x(t')$, which is the unknown you are trying to find. The problem loops back on itself --- exactly the situation Section 4.1 identified as requiring a differential equation.

Layer 4: Debug

Problem 6. A student analyzes a ball launched by a spring. The spring has $k = 500$ N/m and is compressed by 0.08 m. The ball has mass 0.2 kg.

The student writes: "The spring force at the starting position is $F = kx = 500 \times 0.08 = 40$ N. The acceleration is $a = F/m = 40/0.2 = 200$ m/s$^2$. Using $v^2 = v_0^2 + 2a\Delta x$ with $v_0 = 0$ and $\Delta x = 0.08$ m, the launch speed is $v = \sqrt{2(200)(0.08)} = 5.66$ m/s."

(a) What is wrong with this calculation?

(b) Is the student's answer too high, too low, or is it impossible to tell without doing the full calculation?

Check your answer **(a)** The student computed the force at the starting position (maximum compression) and then used it as though it were constant for the entire motion. But the spring force is $F = kx$ --- it depends on how much the spring is compressed. As the ball moves forward and the spring extends, the force *decreases*. By the time the ball separates from the spring, the force is zero. Using the maximum force as though it were constant throughout the motion overestimates the acceleration at every point except the very start. The constant-acceleration formula is invalid here because $a$ is not constant. **(b)** The student's answer is **too high.** The actual acceleration starts at 200 m/s$^2$ and decreases to 0 as the ball moves through the 0.08 m of spring travel. The average acceleration is less than 200 m/s$^2$, so the actual launch speed is less than 5.66 m/s. (The correct launch speed can be found using energy conservation --- a method you will learn in Chapter 7. The answer is $v = \sqrt{k x^2 / m} = \sqrt{500 \times 0.08^2 / 0.2} = 4.0$ m/s, which is indeed less than 5.66 m/s.)

Multiple Representations: The Same Oscillation, Four Ways

To solidify the idea that acceleration is state-dependent, look at a single oscillation of a mass on a spring through four lenses.

Algebraically: $a(x) = -(k/m)x$. The acceleration is a linear function of position. It changes sign as the mass crosses equilibrium.

Graphically ($a$ vs. $x$): A straight line through the origin with negative slope. Every point on this line represents the acceleration at a particular position. The mass traces out this line as it oscillates, moving back and forth along it.

Graphically ($a$ vs. $t$): A sinusoidal curve. The acceleration oscillates in time, reaching its extremes at the turning points and crossing zero at equilibrium. This graph looks like $x(t)$ flipped upside down --- because $a = -(k/m)x$, the acceleration is always proportional to $-x$.

Physically: At the endpoints, the spring is maximally stretched or compressed, the force is at its strongest, and the mass is momentarily at rest. At the center, the spring is relaxed, the force is zero, and the mass is at maximum speed. The mass is always being pulled back toward the center. The acceleration is large where the velocity is zero, and zero where the velocity is large. Force and motion are out of phase.

These are not four separate facts. They are four views of one idea: the acceleration is determined by the current position, and the current position changes as the object moves.

Reflection

Think about the problems you have worked through in this section.

What is the most important thing to check before assuming acceleration is constant?

The answer is simple: check whether the net force depends on position, velocity, or any other quantity that changes during the motion. If it does, the acceleration changes too, and constant-acceleration formulas will give wrong answers.

This check takes five seconds. Look at the forces. Ask: does any of them depend on where the object is or how fast it is going? If the answer is yes --- a spring, a drag force, a changing gravitational field, a changing mass --- then you are in the territory of state-dependent acceleration, and you need the tools of Chapter 4 (differential equations) or Chapter 7 (energy methods) instead of the kinematic equations from Chapter 2.

The conceptual shift is this: acceleration is not an input to the problem. It is an output --- produced by the forces at each moment, changing as the state changes.

Looking Ahead

You have now seen that the "nice" problems of Chapter 2, where acceleration was constant, are the exception rather than the rule. In most real situations, the force depends on the state, the acceleration varies, and the simple kinematic formulas do not apply.

In the next section --- Section 6.5 --- we shift from the forces themselves to the setup of the problem. You have learned how to handle inclines, friction, drag, and nonconstant forces. But each of these problems also required a choice of coordinate system. Sometimes that choice made the algebra clean; sometimes it made it painful. Section 6.5 makes that strategic decision explicit: how do you choose coordinates to simplify the equations of motion?