A heavy chain with a mass per unit length $ρ$ is pulled by the constant force F along a horizontal surface consisting of a smooth section and a rough section. The chain is initially at rest on the rough surface with $x = 0$ . If the coefficient of kinetic friction between the chain and the rough surface is $μ_{k}$ , determine the velocity $v$ of the chain when $x = L$ . The force $F$ is greater than $f_{l i m i t i n g}$ in order to initiate the motion.

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$\sqrt{\frac{4 F}{ρ} - μ_{k} g L}$

$\sqrt{\frac{5 F}{ρ} - μ_{k} g L}$

$\sqrt{\frac{7 F}{ρ} - μ_{k} g L}$

$\sqrt{\frac{2 F}{ρ} - μ_{k} g L}$

answer is A.

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Detailed Solution

$v \cdot \frac{d v}{d x} = \frac{Net force}{mass} = \frac{F - μ_{k} ρ (L - x) g}{ρ L} ∴ \int_{0}^{'} v d v = \int_{0}^{L} \frac{F - μ_{k} ρ (L - x) g}{ρ L} d x ∴ \frac{v^{2}}{2} = \frac{F}{ρ} - μ_{k} g L + \frac{μ_{t} g L}{2} v = \sqrt{\frac{2 F}{ρ} - μ_{k} g L}$