Spring-Launched Block: Does the Detector Get Reached, and Does Mass Matter?
A 0.50 kg block is pressed against a spring of stiffness 200 N/m, compressing it 0.20 m from equilibrium on a horizontal frictionless track. The block is released from rest and later climbs a frictionless incline. A detector is mounted 1.0 m above the base of the incline. In a second trial the incline is unchanged, but a 2.0 m long patch of the horizontal track immediately after the spring has a coefficient of kinetic friction of 0.20. A third trial repeats the second setup with a 1.0 kg block. The group must decide whether the detector is reached in each trial and explain what changes when the mass is doubled. Use g = 9.8 m/s^2 and treat the spring as massless.
1. For the frictionless trial, find the block's speed after leaving the spring and the maximum height it reaches. Is the detector reached?
Read the worked answer
Energy stored in the spring
- U = 1/2 k x^2 = 0.5 x 200 x (0.20)^2 = 4.0 J
Speed after launch (all energy kinetic)
- 1/2 m v^2 = 4.0 J -> v = sqrt(2 x 4.0 / 0.50) = sqrt(16) = 4.0 m/s
Maximum height on the incline
- m g h = 4.0 J -> h = 4.0 / (0.50 x 9.8) = 0.82 m
Verdict 0.82 m < 1.0 m, so the block does not reach the detector; it turns around about 18 cm short. At the top of its path the speed is momentarily zero, and all 4.0 J sits as gravitational potential energy.
2. For the second trial (same 0.50 kg block, 2.0 m rough patch, mu_k = 0.20), find the friction work, the remaining mechanical energy, and the new maximum height. What fraction of the spring energy was dissipated?
Read the worked answer
Friction force and work
- f_k = mu_k m g = 0.20 x 0.50 x 9.8 = 0.98 N
- W_fric = f_k d = 0.98 x 2.0 = 1.96 J (magnitude; energy removed)
Remaining energy available for height
- E_remaining = 4.0 - 1.96 = 2.04 J
- h = 2.04 / (0.50 x 9.8) = 0.42 m
Fraction lost
- 1.96 / 4.0 = 49% of the stored spring energy is dissipated as thermal energy before the incline.
Verdict The block now rises only ~0.42 m, still far short of the 1.0 m detector. Note the friction work is path-dependent and fixed by the 2.0 m patch, so it competes directly with the fixed 4.0 J input.
3. In the third trial the block is 1.0 kg rather than 0.50 kg (same spring compression and same rough patch). Predict the maximum height and explain why mass now matters even though it did not matter in the frictionless trial.
Read the worked answer
Frictionless case (for contrast)
- h = U/(m g): doubling m halves the height per joule, but m also appears in the weight, so h = U/(m g) is unchanged: still 0.82 m. The speed does drop: v = sqrt(2U/m) = sqrt(8.0) = 2.8 m/s.
With friction
- W_fric = mu_k m g d = 0.20 x 1.0 x 9.8 x 2.0 = 3.92 J
- Remaining: 4.0 - 3.92 = 0.08 J
- h = 0.08 / (1.0 x 9.8) = ~0.008 m (about 1 cm)
Equivalently: h = U/(m g) - mu_k d = 0.41 - 0.40 = 0.01 m.
Why the difference The spring input is fixed (4.0 J) and independent of mass, but both the gravitational cost per meter (m g) and the friction cost (mu_k m g d) scale with mass. When the only loss is gravity, mass cancels. When a mass-proportional loss is present, heavier blocks lose a larger fraction of a fixed energy budget, so mass now matters. Assumption checks: block stays on the track, no slipping, spring massless, friction acts only over the 2.0 m patch.