Answers (using g = 9.80 m/s²):
Kinetic energy at the bottom of the first hill (ignore friction):
233 848 J (or 2.34 × 10⁵ J)
Kinetic energy at the top of hill B (31 m tall):
159 113 J (or 1.59 × 10⁵ J)
Speed at the top of hill B:
36.0 m/s
Fastest speed during the entire trip:
43.6 m/s
(This occurs at the bottom of the first hill—or any other bottom point on the track—where all the initial potential energy has been converted to kinetic energy.)
How far the spring would compress (no friction):
75.5 m
Energy lost to friction over the whole ride:
175 386 J (or 1.75 × 10⁵ J)
Step-by-step reasoning (conservation of mechanical energy, no friction until the last part):
Total mechanical energy at the start (top of hill A):
E = m g h_A = 246 kg × 9.80 m/s² × 97 m = 233 848 J
1. Bottom of first hill (h = 0 m):
All PE → KE
KE = E = 233 848 J
2. Top of hill B (h = 31 m):
KE = E – m g h_B = 233 848 J – (246 × 9.80 × 31) = 159 113 J
3. Speed at top of hill B:
KE = ½ m v² → v = √(2 × KE / m) = √(2 × 159 113 / 246) = 36.0 m/s
4. Fastest speed:
Maximum KE occurs at the lowest point (h = 0 m) anywhere on the track.
v_max = √(2 g h_A) = √(2 × 9.80 × 97) = 43.6 m/s
5. Spring compression (ideal, no friction):
At the end of hill C the cart is again at h = 0 m with KE = 233 848 J.
This is all absorbed by the spring: ½ k x² = KE
x = √(2 × KE / k) = √(2 × 233 848 / 82) = 75.5 m
6. Actual compression = ½ of calculated value → x_actual = 75.5 / 2 = 37.75 m
Energy actually stored in the spring = ½ k (x_actual)² = ½ × 82 × (37.75)² = 58 462 J
(This is exactly ¼ of the original 233 848 J.)
Therefore, energy lost to friction during the entire ride = total initial energy – energy left in the spring = 233 848 J – 58 462 J = 175 386 J
(or ¾ of the original mechanical energy).
All values are rounded sensibly for the given data (3 significant figures where appropriate). The calculations assume the bottoms of all hills are at the same reference height (h = 0) and that the spring is located at a bottom point after hill C.
