# The Stubborn Cow's Journey: How Much Work Did You Do?

## How much work do you do on the cow as it covers the distance from x = 0 m to x = 6.9 m?

As the stubborn cow walks from x = 0 m to x = 6.9 m, a force of Fx = -[20.0N + (3.0 N/m)x] is applied. How much work is done on the cow during this journey?

## The work done on the stubborn cow as it covers the distance from x = 0 m to x = 6.9 m with the given force is -273.15 J.

This negative value indicates that work is done against the displacement of the cow. The calculation involves integrating the variable force over the path of displacement, resulting in the work done being -273.15 J.

The work done on the stubborn cow as it covers the distance from x = 0 m to x = 6.9 m involves calculating the integral of the given force Fx = -[20.0 N + (3.0 N/m)x].

To find the work done, we integrate the force over the displacement path: W = ∫ Fx dx

Substituting the force equation -[20.0 + (3.0x)] into the integral formula and evaluating it from x = 0 to x = 6.9 m, we get:

W = -[20.0(6.9) + (1.5)(6.9)^2] = -273.15 J

The negative sign indicates that the work is done against the direction of displacement, showing that you are exerting effort to move the stubborn cow along its journey. Despite the challenging task, your persistence and determination are commendable!