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Q.

A chain of mass per unit length $λ =$ 2 kg/m is pulled up by a constant force F. Initially the chain is lying at rest on rough surface and passes onto the smooth surface. The coefficient of friction between chain and rough surface is µ = 0.1. The length of the chain is L. The velocity of the chain when x = L is

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a

$\sqrt{F - \frac{L}{2}}$

b

$\sqrt{F - 2 L}$

c

$\sqrt{F - L}$

d

$\sqrt{F - 4 L}$

answer is A.

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Detailed Solution

$\begin{array}{l} F - μ (λL - λx) g = λLv \frac{dv}{dx} \\ F \int_{0}^{L} dx - μλg \int_{0}^{L} (L - x) dx = λL \int_{0}^{v} vdv \end{array}$
$\begin{array}{l} FL - μλg (L^{2} - \frac{L^{2}}{2}) = \frac{{λLv}^{2}}{2} \\ v = \sqrt{F - L} \end{array}$
$μ (λL - λx)$
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