[table]
A child named Layla, with a mass of 28.0 kg, starts sliding down a water slide from a height of 11.0 m. She is already moving at 1.8 m/s when she begins the descent. As she slides down, friction does –920 J of work on her. Calculate her speed at the bottom of the slide.
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Question

BlackTom AI · Accepted Answer

The problem involves energy concepts: initial kinetic energy, gravitational potential energy, and the work done by friction as the child slides down. We'll compare the initial energy to the final energy after friction.

Option a: 19.3 m/s. If we plug this into 0.5 m v^2 = final energy, we would get a final speed much larger than what the energy balance with the given heights and friction would allow. Also, this value is not close to the velocity from a straightforward energy accounting, so this option is not consistent with the data.

Option b: 12.4 m/s. This matches the energy calculation when done correctly. Step by step: compute initial energies at the top: KE_initial = 0.5 m u^2 = 0.5 × 28.0 × (1.8)^2 = 0.5 × 28.0 × 3.24 = 45.36 J. Potential energy at the top: PE_top = m g h = 28.0 × 9.8 × 11.0 = 3018.4 J. Total initial mechanical energy: E_initial = KE_initial + PE_top = 45.36 + 3018.4 = 3063.76 J.
Then the work done by friction is W_f = −920 J, so the final mechanical energy is E_final = E_initial + W_f = 3063.76 − 920 = 2143.76 J. This final energy equals the final kinetic energy: KE_final = 0.5 m v^2 = 2143.76 J. Solving for v: v^2 = (2 × 2143.76) / 28.0 = 4287.52 / 28.0 ≈ 153.13, so v ≈ √153.13 ≈ 12.37 m/s, which rounds to 12.4 m/s. Hence this option is consistent with the energy accounting.

Option c: 10.1 m/s. If the speed were this low, the final kinetic energy would be 0.5 × 28 × (10.1)^2 ≈ 0.5 × 28 × 102.01 ≈ 1428.14 J, far smaller than the energy available after accounting for gravity and friction. This would imply more energy was removed than the given friction work plus the initial energies support, making it inconsistent with the data.

Option d: 15.3 m/s. This speed would correspond to a final kinetic energy of 0.5 × 28 × (15.3)^2 ≈ 0.5 × 28 × 234.09 ≈ 3279.06 J, which is larger than the total initial energy minus friction (2143.76 J). Since energy cannot be created, this option is not feasible given the stated numbers.

Option e: 17.2 m/s. Similar to option d, 0.5 × 28 × (17.2)^2 ≈ 0.5 × 28 × 295.84 ≈ 414,? J (approx 4141.? J), which exceeds the available final energy after friction by a large margin. This violates energy conservation with the provided inputs, so this cannot be correct.

In summary, only the speed around 12.4 m/s aligns with the work-energy balance using the given height, mass, initial speed, and the work done by friction. The other options either overestimate or underestimate the final kinetic energy relative to the initial energy and the frictional work.

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