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 this force on an object that moves from xi = 5.41 m to xf = 21.9 m? Answer in units of kJ.

Answer 1

To determine the work done by a varying force, we can integrate the force equation over the given displacement. In this case, the force is given by F = k1x^n - k2. Integrating the force equation from xi = 5.41 m to xf = 21.9 m will give us the work done by this force on the object.

Step-by-step explanation:

The work done by a varying force can be determined by integrating the force with respect to position. In this case, the force is given by F = k1x^n - k2, where n = 3, k1 = 3.6 N/m^3, and k2 = 76 N.

To find the work, we can use the work-energy principle which states that the work done on an object is equal to the change in its kinetic energy. We can calculate the work done by integrating the force equation over the given displacement:

Work = ∫(F(x)dx) = ∫(k1x^n - k2)dx = ∫(3.6x^3 - 76)dx

Integrating this equation from xi = 5.41 m to xf = 21.9 m will give us the work done by the force on the object. The result can be converted to kJ by dividing by 1000.

Answer 2

The work done is 205 kJ.

Step-by-step 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