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Physics Question: In an experiment, one of the forces exerted on a proton is F =−αx^2(vec i), where α=12N/m^2.?
F = -12*x^2 i + 0 j
How much work does vector F do when the proton moves along the straight-line path from the point (0.10m,0) to the point (0.10m,0.40m) ?
The force is only applied in the x direction (i) and the proton moved in the y direction (j). The force did no work. <—
How much work does vector F do when the proton moves along the straight-line path from the point (0.10,0) to (0.3,0)?
Work = integral over the path(F*dx)
Work = integral over the path(-12*x^2 dx) ] from 0.1 to 0.3
Work = -4*x^3 ] 0.1 to 0.3
Work = -4*(.3^3 – .1^3) = -0.104 J
How much work does vector F do when the proton moves along the straight-line path from the point
(0.3, 0) to (0.1,0)?
Opposite the previous answer
Work = 0.104 J
For the 2nd part, the force acts in the negative x direction and the movement is in the positive x direction, therefore energy is used from the system to work past the force, hence the negative sign.
In the 3rd part, the movement is now in the negative x direction, so the force adds energy to the system, hence the positive sign.
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