Ask Question
29 July, 07:59

An applied force varies with position according to F = k1 x n - k2, where n = 3, k1 = 3.6 N/m3, and k2 = 76 N. How much work is done by this force on an object that moves from xi = 5.41 m to xf = 21.9 m? Answer in units of kJ.

+5
Answers (1)
  1. 29 July, 11:04
    0
    The work done is 205 kJ.

    Explanation:

    Hi there!

    Work can be calculated using the following equation:

    W = F · Δx

    Where:

    W = work

    F = applied force

    Δx = displacement

    In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:

    W = ∫ F · dx

    F = 3.6 N/m³ · x³ - 76 N

    W = ∫ (3.6 x³ - 76) dx

    W = 0.9 x⁴ - 76x

    Evaluating from xi to xf:

    W = 0.9 N/m³ (21.9 m) ⁴ - 76 N · 21.9 m - 0.9 N/m³ (5.41 m) ⁴ + 76 N · 5.41 m

    W = 205 kJ
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “An applied force varies with position according to F = k1 x n - k2, where n = 3, k1 = 3.6 N/m3, and k2 = 76 N. How much work is done by ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers