A particle of mass M moves along a horizontal x axis from x=0 to x=L. The coefficient of kinetic friction varies as a function of x as _k(x)= _0- x, where _0, are constants of appropriate dimensions, so that _k(L)=0. The total work done by the frictional force during the motion is n _0 MgL, where g is the acceleration due to gravity. The value of n is :
A. 1 2
Given the coefficient of kinetic friction _k(x) = _0 - x At x = L, _k(L) = 0 _0 - L = 0 = _0 L The frictional force acting on the particle is f_k = _k(x) Mg = ( _0 - _0 L x )Mg The magnitude of work done by the frictional force is given by: |W| = _ 0 ^ L f_k dx |W| = _ 0 ^ L ( _0 - _0 L x ) Mg dx |W| = Mg [ _0 x - _0 x^2 2L ]_ 0 ^ L |W| = Mg ( _0 L - _0 L^2 2L ) = 1 2 _0 MgL Comparing this with n _0 MgL, we get n = 1 2 Answer: 1 2
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