A rope of length L has mass per unit length y varying according to the function Y(x)= e^x/L . The rope is pulled by a force of constant 1N on a smooth horizontal surface. Find the tension in the rope at x= L/2.

Dear Ayush, 

total mass of rope=∫e^(x/L)dx (from x=0 to L) =L(e-1), acceleration =a =force/mass=1/L(e-1), tension at x=L/2 is equal to

 a∫e^(x/L)dx (from x= L/2 to L)= √e/(√e +1)

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