A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x axis from x = 1.40 m to x = 7.00 m, calculate the following.

Your question is incomplete, but i assume they want you to calculate the work performed on the particle as it moves through that distance. That would be the integral of Fx dx from x=1.4 to 7.0

A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x axis from x = 1.80 m to x = 5.40 m, calculate the following.

To calculate the following quantities, we need to use the equation provided for the conservative force, Fx = (2x + 4) N. We also need to determine the work done by the force and the potential energy at both x = 1.40 m and x = 7.00 m.

1. Work done by the force:
The work done by a conservative force as an object moves from one position to another is given by the negative change in potential energy. Therefore, we need to calculate the potential energy at each position and find the difference between them.

To find the potential energy at a given position, we integrate the force equation with respect to x:
U(x) = ∫ Fx dx
= ∫ (2x + 4) dx
= x^2 + 4x + C

To determine the constant of integration (C), we use the initial position x = 1.40 m, where the potential energy is zero:
U(1.40) = (1.40)^2 + 4(1.40) + C
= 1.96 + 5.60 + C
= 7.56 + C
Since the potential energy is zero at this position, C = -7.56.

Now, we can find the potential energy at x = 7.00 m:
U(7.00) = (7.00)^2 + 4(7.00) - 7.56
= 49 + 28 - 7.56
= 69.44 J

Therefore, the work done by the force is the negative change in potential energy:
W = - (U(7.00) - U(1.40))
= - (69.44 - 0)
= - 69.44 J

2. Potential energy at x = 1.40 m:
U(1.40) = 0 J (as calculated earlier)

3. Potential energy at x = 7.00 m:
U(7.00) = 69.44 J (as calculated earlier)