Test a friction-limited stopping prediction

A requested target is a hypothesis

The position control asks whether the initial block can reach a chosen forward position under the selected resistance. It does not place the block there by an external hand. The block starts at x = 0 with speed +4 m/s and kinetic energy 16 J. A target can be reached only if the physical journey extends that far.

Select resistance 6 N. At 2.5 m, friction has removed 15 J and 1 J remains kinetic. For the fixed 2 kg mass, speed is √(2 × 1/2) = 1 m/s. At a 3 m target, the formal energy expression gives 16 − 18 = −2 J. Negative translational kinetic energy is impossible, so this is evidence against reaching the target under the stated initial conditions.

The correct response is to solve for the earlier stop, not to clamp the target’s energy to zero and pretend the block arrived there. The lab displays the stop’s physical ledger but explicitly leaves target speed and target time marked Unreachable. Save both target requests and explain the distinction in your notebook.

Solve the stop using energy

At a stop, final kinetic energy is zero. For a positive constant resistance f, 0 = Kinitial − f dstop, so dstop = Kinitial/f. The units are joules divided by newtons, which are metres. With Kinitial = 16 J and f = 6 N, dstop = 16/6 ≈ 2.667 m.

For 4 N resistance, the stop is exactly 4 m. Request that target: it is reached at rest. Then request 5 m: it is unreachable. The actual stopped state is unchanged, but the answer about reaching the requested target has changed. Exactly reaching a point at zero speed and stopping before it are different outcomes.

At 2 N resistance, the stop would occur at 8 m, beyond the widget’s 6 m display range. The ledger can report that analytic stopping prediction, while the graph only shows the first 6 m. A plot boundary is not a physical barrier. With zero resistance there is no finite stop at all; do not divide by zero or invent a maximum distance.

Check the same prediction with forces and time

While sliding right, acceleration is a = −f/m. For m = 2 kg and f = 6 N, a = −3 m/s². Starting at +4 m/s, velocity reaches zero after tstop = 4/3 ≈ 1.333 s. The average velocity during this constant-acceleration interval is (4 + 0)/2 = 2 m/s. Multiplying by time gives 2 × 4/3 = 8/3 m, matching the energy result.

For a reached intermediate target, the lab calculates first-arrival time from distance divided by average velocity: t = 2d/(vinitial + vfinal). This expression is valid here because acceleration is constant during the sliding interval. At 2.5 m with 6 N, vfinal = 1 m/s, so t = 5/5 = 1 s. Do not use this average-velocity formula for arbitrary time-varying acceleration without checking its assumptions.

The force and energy methods should agree when they describe the same initial state and same physical model. Agreement is a numerical consistency check, not independent experimental validation. Testing real contact would require measured positions or speeds and uncertainties.

Why the block does not reverse

The earlier constant-force lesson kept a signed force acting after velocity reached zero. That model could reverse the particle. Sliding friction has a different contract: it opposes relative sliding and must be reconsidered when sliding ends.

Here no other horizontal force acts. Once the block is at rest relative to the track, the required static friction is zero. The block stays at rest. Continuing to apply −6 N would accelerate it left; that would make the supposed friction point in the same direction as the new sliding motion and create kinetic energy without an energy source. It is an invalid extension of the sliding rule.

The displayed current horizontal force at a stopped state is therefore zero. The deceleration immediately before stopping was still −f/m; the ideal contact model switches regimes at rest. Real transitions can be more complicated, but numerical time stepping should not invent backward motion merely by overshooting zero velocity.

Compare preparations fairly

At fixed initial speed and mass, doubling positive resistance halves both stopping distance and stopping time. Try 4 N and 8 N: the stops are 4 m and 2 m, reached in 2 s and 1 s. Both runs ultimately convert the same initial 16 J into combined internal-energy increase. Larger resistance changes where and how quickly the conversion occurs, not the initial energy available.

A separate prediction concerns initial speed. With the same mass and resistance, dstop = mvinitial²/(2f) and tstop = mvinitial/f. Doubling initial speed quadruples distance but only doubles stopping time. These are ideal track predictions, not road-safety stopping distances: reaction time, tire behavior, slopes, drag and changing contact conditions are absent.

Test an unfamiliar stopping case

A 3 kg block starts at 6 m/s against constant 9 N sliding resistance. Predict initial kinetic energy, stopping distance, stopping time and whether it reaches 5 m. Then predict the stop if initial speed is 12 m/s while everything else stays fixed.

Check the friction-stopping transfer

Initial K = ½ × 3 × 6² = 54 J. The stop is 54/9 = 6 m away and takes 3 × 6/9 = 2 s. At 5 m the block still has 9 J, so it reaches the target with speed √(2 × 9/3) = √6 ≈ 2.449 m/s. At 12 m/s the initial energy is 216 J, stopping distance is 24 m and stopping time is 4 s. With no horizontal drive it remains at rest after either stop.

Use the energy ledger and force model together: one determines the available motion energy, while the other supplies direction, timing and the rule change at rest. Neither replaces observation of a real surface.