Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Potential and Kinetic Energy on a Frictionless Track — The Sled Run

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  MID-LEVEL  |  12 points  |  30 min

▤ Scenario

A sled of mass m slides along a frictionless, icy track. From a high platform the track drops into a dip, rises over a hump, drops into a second dip and then rises to a long level stretch. Figure 1 shows the gravitational potential energy U_g of the sled–Earth system as a function of the sled's horizontal position x. Energies are given in multiples of a constant U_0.

The sled is released from rest at position x_1, where U_g = 6U_0. Positions x_2 and x_4 are the bottoms of the two dips, x_3 is the top of the hump, and the level stretch begins at x_5.

Figure 1. Gravitational potential energy of the sled–Earth system as a function of position.

Air resistance is negligible. The sled stays on the track at all times, and the system is the sled and Earth.

✎ Free Response Questions

Part A

The energy bar chart for position x_1 is shown in Figure 2. Complete the energy bar charts in Figure 2 for positions x_2 and x_3 by drawing bars that represent the kinetic energy K of the sled and the gravitational potential energy U_g of the sled–Earth system.

• Each bar must start on the zero line.

• Represent any energy that is equal to zero with a distinct line on the zero line.

Figure 2. Energy bar charts. The chart for x_1 is complete.

Part B

Starting with conservation of energy, derive an expression for the speed v_3 of the sled at position x_3. Express your final answer in terms of m, U_0, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

On the graph in Figure 1, do the following for the motion of the sled from x_1 to the right-hand edge of the graph.

• Sketch and label a line or curve that represents the total mechanical energy E of the sled–Earth system.

• Sketch and label a line or curve that represents the kinetic energy K of the sled.

Part D

The sled is now released from rest at a lower point on the first slope, between x_1 and x_2, where U_g = 3U_0. Use the graph in Figure 1 to predict whether the sled reaches position x_4.

______ The sled reaches x_4.

______ The sled does not reach x_4.

Briefly justify your response by referencing a fundamental physics principle.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board's published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A — Model Answer

From Figure 1, U_g = U_0 at x_2 and U_g = 4U_0 at x_3. The track is frictionless, so the total mechanical energy keeps its release value: E = K + U_g = 0 + 6U_0. The kinetic energy is whatever is left over, K = 5U_0 at x_2 and K = 2U_0 at x_3. In every chart the two bars add up to the same 6U_0.

Figure 3. Completed charts: the two bars total 6U_0 at every position.

Scoring (3 points):

▸ A1 — 1 point: For U_g bars of height U_0 at x_2 and 4U_0 at x_3.

▸ A2 — 1 point: For K and U_g bars whose heights add to the same total at x_2 and at x_3.

▸ A3 — 1 point: For bars whose total at each position is 6U_0, the same as at x_1 (K = 5U_0 at x_2 and K = 2U_0 at x_3).

⚠︎ Common error (partial credit): Correct U_g bars with K bars that do not bring the totals to a common value — earns 1 of 3 points (A1 only).

✗ Common error (no credit): Drawing the U_g bar at 6U_0 in all three charts. U_g is read from the curve at each position; it is the total that stays the same.

Part B — Model Answer

Mechanical energy of the sled–Earth system is conserved between x_1 and x_3.

K_1 + U_g,1 = K_3 + U_g,3
0 + 6U_0 = ½m(v_3)² + 4U_0
½m(v_3)² = 2U_0
v_3 = 2√(U_0/m)

The speed at the top of the hump depends only on how far the curve has dropped below its release value. Kinetic energy is the gap between the release energy and the curve.

Scoring (4 points):

▸ B1 — 1 point: For a statement or equation of conservation of mechanical energy between x_1 and x_3.

▸ B2 — 1 point: For zero kinetic energy at release and kinetic energy ½m(v_3)² at x_3.

▸ B3 — 1 point: For substituting the potential energies read from Figure 1, 6U_0 at x_1 and 4U_0 at x_3.

▸ B4 — 1 point: For a correct final expression, v_3 = 2√(U_0/m), or an equivalent form such as √(4U_0/m).

Scoring Note: A correct, isolated, final expression for v_3 earns points B3 and B4.

⚠︎ Common error (partial credit): Using the drop from the platform, 7U_0 − 4U_0, in an otherwise correct derivation, giving v_3 = √(6U_0/m) — earns 2 of 4 points (B1 and B2).

✗ Common error (no credit): Setting ½m(v_3)² = 4U_0, the potential energy at x_3. This equates kinetic energy to the height of the curve instead of the gap above it.

Part C — Model Answer

E is a horizontal line at 6U_0, starting at x_1. At every position K = E − U_g, so the K curve is the U_g curve turned upside down: zero at x_1, a maximum of 5U_0 at x_2, a dip to 2U_0 at x_3, a second peak of 4U_0 at x_4, and constant at 2U_0 beyond x_5, where the sled moves at constant speed. Wherever U_g is lowest, K is highest.

Figure 4. E is constant at 6U_0; K is the gap between E and the U_g curve.

Scoring (3 points):

▸ C1 — 1 point: For a horizontal line at 6U_0 that begins at x_1, with a label.

▸ C2 — 1 point: For a labelled K curve that starts at zero at x_1, has maxima at x_2 and x_4 with a minimum between them at x_3, and is horizontal beyond x_5.

▸ C3 — 1 point: For K values consistent with K = E − U_g: 5U_0 at x_2, 2U_0 at x_3, 4U_0 at x_4 and 2U_0 beyond x_5, each within half a division.

Scoring Note: Point C3 may be earned for values consistent with the E line drawn for point C1.

⚠︎ Common error (partial credit): A correct E line and a K curve of the right shape with wrong values, such as a peak of 6U_0 at x_2 — earns 2 of 3 points (C1 and C2).

✗ Common error (no credit): Drawing K with the same shape as U_g, high on the hump and low in the dips. The sled is slowest at the top of the hump.

Part D — Model Answer

The sled does not reach x_4.

Released from rest where U_g = 3U_0, the system has total mechanical energy E = 3U_0, and that total is conserved. At x_3 the curve is at 4U_0, above E. Reaching x_3 would need K = 3U_0 − 4U_0 = −U_0, and kinetic energy cannot be negative. The sled passes x_2, climbs until the curve reaches 3U_0 between x_2 and x_3, stops for an instant and slides back. A position can be reached only if the U_g curve stays at or below the E line all the way there, even though x_4 itself is lower than the release point.

Scoring (2 points):

▸ D1 — 1 point: For indicating that the sled does not reach x_4.

▸ D2 — 1 point: For a justification based on conservation of energy: the total mechanical energy, 3U_0, is less than U_g at x_3 (4U_0), so the sled cannot pass the hump.

✗ Common error (no credit): Selecting "reaches x_4" because U_g at x_4 (2U_0) is lower than at the release point (3U_0). This compares only the start and end points and ignores the hump between them.

Keep going:

Equations & Traps deck · Mind Maps bundle