Question

A force given by F(x) = 5x³(in n/m³) acts on a 1-kg mass moving on a frictionless surface. The mass moves from x = 3.67 m to x = 6.33 m. How much work is done by the force?

Answer 1

The work done by the force F(x) = 5x³ on a 1-kg mass as it moves from x = 3.67 m to x = 6.33 m is calculated using integration, resulting in approximately 658.08 Joules of work.

Step-by-step explanation:

The student asks how much work is done by a force, represented by the function F(x) = 5x³, on a 1-kg mass as it moves from x = 3.67 m to x = 6.33 m. To calculate the work done by a variable force along a straight path, we need to integrate the force function with respect to displacement. This is a concept from physics that involves some mathematics, specifically calculus.

Let's carry out the integration:

1. Firstly, set up the integral of the force function F(x) with respect to x from 3.67 to 6.33 m.
2. Calculating the integral, ∫ F(x) dx = ∫ 5x³ dx, we have: ∫ 5x³ dx = frac{5}{4}x^4 |_{3.67}^{6.33}
3. Evaluate the definite integral: frac{5}{4}(6.33)^4 - frac{5}{4}(3.67)^4 = 658.08 J

Therefore, the work done by the force as the mass moves from x = 3.67 m to x = 6.33 m is approximately 658.08 Joules.