Final answer:
The work done by the force F(x) = (-5.0x² + 7.0x) N on a particle moving from x = 2.0 m to x = 5.0 m is calculated using integration. The correct work found after performing the integral and substituting the limits is 1.75 J, which doesn't match the provided answer choices.
Step-by-step explanation:
The question is asking to calculate the work done by a variable force as it acts on a particle moving from one position to another along the x-axis. The force in question is described by the equation F(x) = (-5.0x² + 7.0x) N. The work done by this force as the particle moves from x = 2.0 m to x = 5.0 m can be found using the work integral:
W = ∫ F(x) dx, from x = 2.0 m to x = 5.0 m
This integral can be calculated by integrating the force function F(x) over the given limits. The integration of the function F(x) = (-5.0x² + 7.0x) yields a new function representing the work done:
W = [-5.0x³/3 + 7.0x²/2] from 2.0 m to 5.0 m
Plugging in the limits gives:
W = [-5.0(5.0)³/3 + 7.0(5.0)²/2] - [-5.0(2.0)³/3 + 7.0(2.0)²/2]
W = [-125.0/3 + 87.5/2] - [-20.0/3 + 14.0/2]
W = (-41.67 + 43.75) - (-6.67 + 7.00)
W = 2.08 - 0.33 = 1.75 J
Therefore, the total work done by the force as the particle moves from x = 2.0 m to x = 5.0 m is 1.75 J, which is not one of the provided answers. It is possible that there may be an error in the provided options or a mistake in the calculation. Based on the given equation for the force and the process used to integrate it, the correct answer can be determined by recalculating the work done.