A hiker pulls a 12.0 kg sled up a snowy incline 5.00 m long and 30.0° above horizontal. If the sled moves at constant speed, how much work does the hiker do against gravity alone?

Question

BlackTom AI · Accepted Answer

To determine the work done against gravity, focus on the vertical rise of the sled, since gravity affects vertical height, not horizontal distance.
Option A: 294 J. The vertical height gained is h = (5.00 m) × sin(30°) = 5.00 × 0.5 = 2.50 m. The gravitational potential energy change is ΔU = m g h = (12.0 kg) × (9.8 m/s²) × (2.50 m) = 12 × 9.8 × 2.5 = 12 × 24.5 = 294 J. This matches the given value, so this option is consistent with Wgrav = m g h.
Option B: 300 J. This is close to 294 J but not exact. The small difference could come from rounding g, but using standard g = 9.8 m/s² with the exact height (2.50 m) yields 294 J. Since the problem typically uses 9.8 m/s², 300 J would be inaccurate.
Option C: 510 J. This would correspond to a larger height or mass (e.g., if h ≈ 4.25 m with m ≈ 12 kg: 12 × 9.8 × 4.25 ≈ 499 J). Here the vertical rise is only 2.50 m, so 510 J is too large.
Option D: 600 J. This implies a height around h ≈ 5.0 m, since 12 × 9.8 × h ≈ 600 J leads to h ≈ 600/(12×9.8) ≈ 5.1 m. Again, the incline height is only 2.50 m, so this value overestimates the work.
Overall, the work against gravity equals m g h with the computed height 2.50 m, yielding 294 J, while the other options either overestimate or misrepresent the vertical rise involved in the motion.

SPH4U - Physics 12 (2025-26) - A

A hiker pulls a 12.0 kg sled up a snowy incline 5.00 m long and 30.0° above horizontal. If the sled moves at constant speed, how much work does the hiker do against gravity alone?

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