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

Solution 60P x 2 Work done by a variable force is given by, W = Fdx…..(1) , where F is the force that x 1 is a function of x, x a1d x are 2nitial and final displacements. Given that, Fx = -[20.0 N + (3.0 N/m)x] x = 0, x = 6.9 m 1 2 Substituting these values in equation 1, 6.9 W = [20.0 N + (3.0 N/m)x]dx 0 6.9 W = (20.0dx + 3.0xdx) 0 6.9 6.9 W = 20.0 dx 3.0xdx 0 0 (6.9) W = 20.0 × (6.9 0) 3.0 × 2 J W = 138 J 71.4 J W = 209.4 J W 209 J Thus, the force applied by you does a work of approximately -209 J.