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

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