Spring Launcher on a Friction Incline: Model versus Measurement
Students launch a 0.40 kg cart from rest up a straight track inclined at 25° to the horizontal. A spring of spring constant 180 N/m is compressed by 0.12 m and released; the cart leaves the spring at the bottom of the incline and slides upward. The coefficient of kinetic friction between cart and track is 0.15, and the coefficient of static friction is 0.25. Ignoring friction, a student predicts a vertical rise of 0.33 m. In the first experiment the cart travels 0.59 m along the incline before stopping momentarily; on a later day, when the track was visibly dusty, a group measures 0.47 m. After stopping, the cart either stays where it is or slides back toward the spring. The class must decide whether the first launch is consistent with the friction model, what the dusty-track discrepancy means, and whether the cart will return to the spring.
1. Calculate the distance the cart should travel up the incline before stopping, using energy considerations and including kinetic friction. Compare with the frictionless prediction.
Read the worked answer
Spring energy stored: E = ½kx² = ½ × 180 × (0.12)² = ½ × 180 × 0.0144 = 1.296 J.
Frictionless case: mgh = E, so h = 1.296/(0.40 × 9.8) = 1.296/3.92 = 0.331 m, and the distance along the slope is d = h/sin25° = 0.331/0.4226 = 0.783 m. The student's 0.33 m is a vertical rise, not a distance along the track; confusing the two is the first trap.
With friction, energy leaves the mechanical system as heat. The two forces opposing motion along the slope are the gravity component and kinetic friction: mg sin25° = 0.40 × 9.8 × 0.4226 = 1.657 N μₖmg cos25° = 0.15 × 0.40 × 9.8 × 0.9063 = 0.533 N total = 2.190 N.
Then E = (2.190)d, so d = 1.296/2.190 = 0.592 m.
The model predicts 0.59 m, which matches the first experiment to two significant figures. That agreement supports the friction model and the stated spring constant, and it shows that friction accounts for the shortfall from 0.78 m, not an error in the spring energy.
2. The dusty-track trial gives 0.47 m instead of 0.59 m. Propose an explanation and describe a measurement that would distinguish it from a fault in the spring or cart.
Read the worked answer
From d = E/F_opposing, a shorter distance means either less energy was delivered or the opposing force was larger. For d = 0.47 m with the same 1.296 J, the required opposing force is 1.296/0.47 = 2.76 N, about 0.57 N more than the model's 2.19 N. If that extra force is friction, μ would have risen from 0.15 to roughly (2.76 − 1.657)/(0.40 × 9.8 × 0.9063) ≈ 0.31, which is plausible for a dusty track.
Alternative explanations: the spring may not have been compressed the full 0.12 m, the spring may have lost stiffness, the cart's wheels may bind, or the track may have been tilted differently. These are distinguishable by measurement rather than argument. Measure the cart's speed just after it leaves the spring with a light gate. If v ≈ √(2E/m) = √(2 × 1.296/0.40) = 2.55 m/s, the spring delivered the expected energy and the loss is downstream (friction). If v is clearly lower, the spring or the compression is at fault. Separately, measure μₖ directly by finding the angle at which the cart slides down at constant speed (tan θ = μₖ) on the dusty and cleaned track. Repeating the launch after cleaning the track should restore 0.59 m if dust is the cause.
3. After the cart stops momentarily, will it slide back toward the spring? Justify with a force comparison, and state what happens on later passes.
Read the worked answer
At the top, the component of weight down the slope is mg sin25° = 1.657 N. The maximum static friction available is μₛmg cos25° = 0.25 × 0.40 × 9.8 × 0.9063 = 0.888 N.
Because 1.657 N > 0.888 N, static friction cannot hold the cart and it accelerates back down the slope. The same comparison holds at the moment of stopping; static friction is not large enough to lock the cart in place.
As it descends, kinetic friction acts up the slope (opposing motion), so the net downward force is 1.657 − 0.533 = 1.124 N over the 0.592 m, giving about 0.665 J of kinetic energy at the bottom - well below the 1.296 J originally stored. The cart will compress the spring with roughly half the original energy, rebound, and stop after a shorter distance again. Each pass removes energy through friction, so the motion decays and the cart will not return to the original compression. This also explains why a 'stop' on an incline is only momentary and why energy accounting, not force balance alone, predicts the sequence.