A block of mass 5 kg is pulled along a horizontal surface. The surface roughness varies such that the coefficient of kinetic friction depends on the position x as (x) = 0.03 x^2 , where x is in meters. The work done against friction as the block moves from x = 0 to x = 10 m is: (Take g = 10 m/s ^2 )

Step-by-step solution

Given: Mass of the block, m = 5 kg Coefficient of friction, (x) = 0.03 x^2 Initial position, x₁ = 0 Final position, x₂ = 10 m The frictional force is variable and depends on position: f(x) = (x) m g = (0.03 x^2)(5)(10) = 1.5 x^2 N The work done against friction is the integral of the frictional force over the displacement: W = _ x₁ ^ x₂ f(x) , dx W = ₀¹⁰ 1.5 x^2 , dx W = 1.5 [ x^3 3 ]₀¹⁰ W = 0.5 (10^3 - 0^3) = 0.5 1000 = 500 J Answer: 500 J