Problem Statement
A conveyor belt moves horizontally at speed v₀. A box of mass m is gently placed on the belt. The coefficient of kinetic friction is μk. Find: (a) the time until the box reaches belt speed; (b) the distance the box travels in that time.
Given
v₀, m, μk. Box placed at rest.
Concepts & Formulas
Friction accelerates box: a = μk·g. Time to reach v₀: t = v₀/(μk·g). Distance: s = v₀²/(2μk·g).
Step-by-Step Solution
Step 1: a = μk·g.
Step 2: v = at = μk·g·t = v₀ → t = v₀/(μk·g).
Step 3: s = at²/2 = μk·g·v₀²/(2μk²·g²) = v₀²/(2μk·g).
Worked Calculation
t = v₀/(μk·g). s = v₀²/(2μk·g).
Boxed Answer
t = v₀/(μk·g); s = v₀²/(2μk·g)
Physical Interpretation
The box travels half the distance the belt moves in time t (belt moves v₀t = v₀²/(μk·g)), confirming that kinetic friction does work equal to the heat generated by slipping.
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