Consider the blocks shown below. Assume the rope is massless and the pulley is massless and frictionless. The mass of block M is 4.0 kg and the mass of block m is 2.0 kg. The coefficient of kinetic friction between block M and the table is 0.22. Assuming the blocks start at rest, what is the speed of the blocks when the block m has fallen down a height h = 0.85 m ?

Question

Consider the blocks shown below.    Assume the rope is massless and the pulley is massless and frictionless. The mass of block M is 4.0 kg and the mass of block m is 2.0 kg.  The coefficient of kinetic friction between block M and the table is 0.22.  Assuming the blocks start at rest, what is the speed of the blocks when the block m has fallen down a height h = 0.85 m ?

BlackTom AI · Accepted Answer

We start by identifying the energy transfers in the system as the blocks move a vertical distance h = 0.85 m. Since the rope is ideal (massless, inextensible) and the pulley is massless and frictionless, the horizontal displacement of mass M along the table equals the vertical drop of mass m, so the distance moved along the table is s = h.

Option-by-option analysis:
- 5.5 m/s: If this were the speed, the kinetic energy (1/2)(M+m)v^2 would be much larger than the gravitational potential mgh available after subtracting friction work. This would require a net energy gain rather than the energy balance we expect, making this value inconsistent with the energy conservation including friction losses.
- 1.8 m/s (approximately correct): This value is consistent with the energy balance that accounts for both kinetic energy of both masses and the work done against kinetic friction on M as it slides. Using v^2 = [2 g h (m − μ M)] / (M + m), substituting M = 4 kg, m = 2 kg, μ = 0.22, g = 9.8 m/s^2, h = 0.85 m yields v ≈ 1.76 m/s, which rounds to 1.8 m/s. The expression shows how friction reduces the net energy available for kinetic energy.
- 1.3 m/s: This is smaller than the calculated value, suggesting either a larger friction term or a smaller net gravitational conversion to kinetic energy than actually present. If friction were stronger than μ M with the given M, the result would be lower, but the computation with the provided parameters gives a value near 1.76 m/s, not as low as 1.3 m/s.
- 2.9 m/s: This would require substantially more energy to be converted into kinetic energy than is provided after accounting for friction losses. The required v^2 would be much larger than the computed value, which is inconsistent with the energy balance for the given parameters.
- 0.90 m/s: This speed is noticeably smaller than the actual outcome. It would imply a much larger frictional loss or smaller net gravitational contribution than the parameters produce. The calculation with the given masses and μ shows a higher speed near 1.76–1.8 m/s.

In summary, the correct approach is to apply energy conservation with friction: m g h − μ M g h = (1/2)(M + m) v^2, yielding v^2 = [2 g h (m − μ M)] / (M + m). Substituting the numbers confirms v ≈ 1.76 m/s, which aligns with the 1.8 m/s choice when rounded.

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