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 :

Step-by-step solution

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