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 F _ { x } = - [ 20.0 N + ( 3.0 N / m ) x ]F x =−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

Answer 1

Answer:

The work done is -209.42 J.

Explanation:

F(x) = (- 20 - 3 x ) N

x = 0 to x = 6.9 m

Here, the force is variable in nature, so the work done by the variable force is given by

$W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J$