Question

A uniform chain of length l and mass m overhangs on a rough horizontal table with its 3/4 part on the table. The friction coefficient between the table and the chain is µ. Find the magnitude of work done (in Joule) by the friction during the period the chain slips off the table. (Take m = 0.2, g = 10 m/s2 , L = 2m, m = 16 kg)

Answer 1

A uniform chain of length l and mass m overhangs on a rough horizontal table with its 3/4 part on the table

Consider an element (dx) of mass ( dm )

Let the frictional force acting on this element be

dF

$\implies{dF = - \mu \: dm \: g}$

here ,

the value of dF is negative as the direction of frictional force is opposite to the direction of sliding of the chain

If the chain of mass m has length L then the chain

of mass dm will have length dx

By unitary method we get ,

$\implies{dm = \frac{m}{L} dx}$

Substituting the value of dm

$\implies{dF = - \mu \: \frac{m}{L} dx\: g}$

The work done is given by the formula ,

$\implies{w = - \int{dF \: x}}$

$\implies{w = - \int{ - \mu \: \frac{m}{L} \: g \: x dx}}$

$\implies{w = \mu \: \frac{m}{L} g \: \displaystyle\int\limits_{0}^{ \frac{3}{4} }xdx}$

$\implies{w = \mu \: \frac{m}{L} g\lim_{0 \to \frac{3}{4} } \frac{ {x}^{2} }{2} }$

$\implies{w = \frac{9 \: m \mu g }{32 \: L} }$

given ,

• mass of the chain = 16 kg
• acceleration due to gravity = 10 m/s ²
• coefficient of friction = 0.2
• length of the chain = 2 m

Substituting the above values ,

$\implies{w = \frac{9 \: \times 16 \times 0.2 \times 10}{32 \times 2} }$

$\boxed{w = 4.5J}$