Question

How much work does the force F(x)=(−2.0/x)N do on a particle as it moves from x=2.0m to x=5.0m?

a) -6 J
b) -8 J
c) -10 J
d) -12 J

Answer 1

The work done by the force F(x) = (-2.0/x) N as it moves from x = 2.0 m to x = 5.0 m can be calculated by integrating the force over the path, which gives approximately -1.8 J, meaning that none of the provided options are correct.

Step-by-step explanation:

The student asked how much work the force F(x) = (-2.0/x) N does on a particle as it moves from x = 2.0 m to x = 5.0 m. To find the work done by a variable force, we integrate the force function over the specified distance. The work done, W, can be calculated using the integral of F(x) with respect to x over the limits from 2.0 m to 5.0 m.

W = ∫2.05.0 F(x) dx = ∫2.05.0 (-2.0/x) dx = -2.0 * ln(x) |2.05.0 = -2.0 * (ln(5.0) - ln(2.0)) = -2.0 * ln(5.0/2.0) = -2.0 * ln(2.5)

Plugging in the value for natural log of 2.5, we get:
W = -2.0 * ln(2.5) = -2.0 * 0.9163 = -1.8326 J (approximately)

Therefore, none of the provided options (a, b, c, d) are correct based on the calculation.