CALC In an experiment, one of the forces exerted on a proton is where ? = 12 N/m2. (a) How much work does do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force conservative? Explain. If is conservative, what is the potential-energy function for it? Let U = 0 when x = 0.

Solution 30E x 2 Work done by a variable force, say F(x) is given by, W =F(x)dx…..(1), where x i1 x1 the initial displacement and x is the final displacement of the body. 2 2 Given that, F = x i Given, = 12 N/m 2 2 So, F = 12x i (a) When the body moves from the point (0.10 m,0) to (0.10 m, 0.40 m), there is displacement only along the y-axis, No displacement happens along the x-axis. The force F = 12x i acts along the x-axis. Hence force and displacement are perpendicular to each other. This signifies that work done by the given force for the given displacement is zero, because both are perpendicular to each other. (b) For the displacement (0.10 m,0) to (0.30 m,0), work done by the force is calculated as below. From equation (1), 0.30 W = ( 12x i )dx 1 0.10 x 3 0.30 W 1 = 12i ( 3 0.10...