In coordinates with the origin at the barn door, the cow walks from x 0 to x 6.9 m as you apply a force with x component Fx 320.0 N 13.0 N m2x4. How much work does the force you apply to do on the cow during this displacement?

-209.42J

Explanation:

Here is the complete question.

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=-[20.0N + (3.0N/m) x]. How much work does the force you apply do on the cow during this displacement?

Solution

The work done by a force W = ∫Fdx since our force is variable.

Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx = -[20.0N + (3.0N/m) x] the force applied on the cow.

So, the workdone by the force on the cow is

W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹-[20.0N + (3.0N/m) x] dx

= ∫₀⁶°⁹-[20.0Ndx - ∫₀⁶°⁹ (3.0N/m) x] dx

= - [20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹

= - [20 * 6.9 - 20 * 0] - [3.0 * 6.9²/2 - 3.0 * 0²/2]

= - [138 - 0] - [71.415 - 0] J = (-138 - 71.415) J

= - 209.415 J ≅ - 209.42J