100
The problem relates a locomotive engine's speed (\(S\)) to the number of wagons (\(w\)) attached. The base speed (no wagons) is \(S_0 = 40\) km/hr. The speed reduction (\(\Delta S\)) is directly proportional to the square root of the number of wagons: \(\Delta S = k \sqrt{w}\) where \(k\) is the constant of proportionality.
The speed of the engine with \(w\) wagons is given by: \(S(w) = S_0 - \Delta S\) \(S(w) = 40 - k \sqrt{w}\)
We use the given information: when \(w=36\) wagons, the speed \(S = 16\) km/hr. Substitute these values into the equation: \(16 = 40 - k \sqrt{36}\) \(16 = 40 - k \times 6\)
Solve for \(k\): \(6k = 40 - 16\) \(6k = 24\) \(k = \frac{24}{6} = 4\)
With \(k=4\), the speed equation becomes: \(S(w) = 40 - 4 \sqrt{w}\)
The engine is unable to move when its speed \(S(w)\) is less than or equal to zero. We find the number of wagons where the speed is exactly zero by setting \(S(w) = 0\): \(0 = 40 - 4 \sqrt{w}\)
Solve for \(w\): \(4 \sqrt{w} = 40\) \(\sqrt{w} = \frac{40}{4} = 10\) \(w = 10^2 = 100\)
This means the engine stops moving when 100 wagons are attached. Therefore, the smallest number of wagons with which the engine is unable to move is 100.
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