A 700 lb floating platform is at rest when a 200 lb crate is thrown onto it with a horizontal speed of v0 = 12 ft/s. Once the crate stops sliding relative to the platform, the platform and crate move together with a speed of v = 2.667 ft/s. Neglect the vertical motion of the system and the resistance due to the relative motion of the platform with respect to the water. Determine the distance that the crate slides relative to the platform if the coefficient of kinetic friction between the platform and the crate is 0.25.

To determine the distance that the crate slides relative to the platform, we can use the principle of conservation of linear momentum and the work-energy principle. Here are the steps:

1. First, we need to find the initial velocity of the platform (v_platform_initial). Since the platform is initially at rest, its initial velocity is 0 ft/s.

2. Apply the conservation of linear momentum to the system (crate + platform) before and after the collision:

m_crate * v0 + m_platform * v_platform_initial = (m_crate + m_platform) * v

where m_crate = 200 lb, m_platform = 700 lb, and v = 2.667 ft/s.

3. Solve for the initial velocity of the crate relative to the platform (v_crate_initial_relative):

v_crate_initial_relative = v0 - v = 12 ft/s - 2.667 ft/s = 9.333 ft/s

4. Use the work-energy principle to relate the initial and final kinetic energies of the crate and the work done by friction:

(1/2) * m_crate * v_crate_initial_relative^2 - f_friction * d = 0

where f_friction = μ * m_crate * g, μ = 0.25 (coefficient of kinetic friction), g = 32.2 ft/s^2 (acceleration due to gravity), and d is the distance slid.

5. Solve for the distance (d):

(1/2) * 200 * (9.333)^2 - 0.25 * 200 * 32.2 * d = 0

6. Solve for d:

d = (1/2) * 200 * (9.333)^2 / (0.25 * 200 * 32.2) ≈ 13.49 ft

So the distance that the crate slides relative to the platform is approximately 13.49 ft.

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