Final answer:
The work done on the stubborn cow as it covers the distance from x = 0 m to x = 6.9 m with the given force is -273.15 J, indicating work is done against the displacement.
Step-by-step explanation:
To calculate the work done on the cow as it covers the distance from x = 0 m to x = 6.9 m with the force Fx = -[20.0 N + (3.0 N/m)x], we need to integrate the force over the path of the displacement. Since the force is variable, we can express the work done as an integral:
W = ∫ Fx dx
In this case, the integral would be:
W = ∫ (-[20.0 + (3.0x)]) dx from x = 0 to x = 6.9 m
This evaluates to:
W = -[20.0x + (3.0/2)x^2] from x = 0 to x = 6.9
We plug in the limits of integration and find:
W = -[20.0(6.9) + (1.5)(6.9)^2] = -273.15 J
The negative sign indicates that the work is being done against the direction of displacement, which makes sense because the cow is stubborn and resists moving back to the barn.