Use energy to test a stopping target

A negative result can reject the proposed journey

The particle starts with kinetic energy 16 J and velocity +4 m/s. Select constant force −4 N and request a target at +5 m. Multiplying force by the requested displacement gives a signed area of −20 J. The expression Kinitial + W is then 16 − 20 = −4 J.

This is not negative physical kinetic energy. It says the specified particle cannot reach that forward target under the stated force. The algebra has evaluated a proposed endpoint without establishing that the trajectory passes through it. The interface keeps the negative expression visible as a diagnostic, but assigns no physical energy, speed or arrival time to the unreachable target.

Clamping the result to zero would hide the problem. It would falsely suggest that the particle arrives at +5 m and stops there. Instead, solve for the actual position where its initial kinetic energy has been removed.

Find the forward turning point

For a constant opposing force of magnitude B, work over a forward displacement d is −Bd. The letter B is a positive force magnitude, not the signed force control. At the turning point Kfinal = 0, so 0 = Kinitial − Bdstop, giving dstop = Kinitial/B.

With Kinitial = 16 J and B = 4 N, dstop = 16/4 = 4 m. At +3 m the work is −12 J and energy remains 4 J, giving speed 2 m/s. At +4 m the work is −16 J and speed reaches zero. At +5 m the target is beyond the turn and is not reached.

The force-position plot may shade the formal area to a requested target beyond the turning point. The short-dashed line identifies the turn when it falls within the plot; the long-dashed line identifies the requested target. The warning matters: the entire shaded area is not work actually accumulated by the moving particle. Its outward journey ends at 4 m after only −16 J of work.

Try −6 N. The turning position is 16/6 = 2.666… m, between the available integer target settings. A target of 2 m is reachable and a target of 3 m is not. The model calculates the turning point analytically; it does not round the stopping position to the nearest control step or plotted tick.

Check the result against motion equations

The mass is 2 kg, so force −4 N gives acceleration a = −2 m/s². Starting from +4 m/s, the velocity reaches zero after tstop = 4/2 = 2 s. Its displacement is 4 × 2 + ½(−2) × 2² = 4 m, agreeing with the energy calculation.

For a reachable forward target, the widget finds first-arrival time from the average velocity under constant acceleration: displacement equals (initial velocity + final velocity) × time / 2. Here both endpoint velocities on that first forward segment are nonnegative, so time is 2d/(vinitial + vfinal). This form also handles zero force and the exact turning point without dividing by acceleration.

At target zero, first arrival is time zero. The model does not silently select a later return to the origin. At an unreachable target it reports no arrival time; a numerical square root of a negative energy is not repaired into a plausible-looking duration.

Zero velocity does not mean the force switches off

The selected negative force remains negative. At the turn, velocity is momentarily zero but acceleration is still −2 m/s². The particle then reverses and moves left. After reversal, both force and displacement over a short interval are negative, so their product is positive: the same force now adds kinetic energy.

This is different from friction that changes direction with relative sliding, and different from a controller that reduces force and holds an object stopped. The earlier friction lesson shows that the contact regime must be reconsidered when velocity reaches zero. Our constant-force model deliberately does not claim to implement either mechanism.

At +3 m, the particle has speed 2 m/s on its first forward arrival and again on its later backward passage under the continued constant force. Kinetic energy is 4 J at both passages, but velocity is +2 m/s first and −2 m/s later. An energy equation can identify speed at a position without determining which passage or direction is intended.

Use stopping estimates within their assumptions

Because Kinitial = ½mv², dstop = mv²/(2B) under the constant opposing-force assumption. At fixed mass and fixed B, doubling initial speed quadruples the stopping distance. The relationship does not mean every real brake produces constant force, nor does it include a delay before braking begins.

For a changed example, a 2 kg particle starting at 8 m/s has kinetic energy 64 J. A constant 4 N opposing force requires 16 m to remove it, compared with 4 m at 4 m/s. That higher-speed example lies outside this widget’s fixed initial condition and is a worked calculation, not an extra selectable experiment.

The OpenStax work-energy discussion provides the general accounting relationship. Real stopping evidence requires measured forces or justified force models, actual initial conditions and uncertainty. These model results do not measure physical braking performance or establish a safe operating distance.

Transfer: A 1 kg cart starts at 6 m/s. A constant −3 N force acts on a horizontal frictionless track. Is +5 m reachable? What is the stopping position, and what does the continued force do afterward?

Check the stopping-energy transfer

Initial kinetic energy is ½ × 1 × 6² = 18 J. Work to +5 m is −15 J, leaving 3 J, so that target is reached with speed √6 ≈ 2.449 m/s on the first forward passage. The stopping position is 18/3 = 6 m. Continued negative force then reverses the cart; it does not hold the cart at +6 m.