A constant force F has magnitude 30 Newtons and the same dir

A constant force F has magnitude 30 Newtons and the same direction as the vector a = 3I - 4 J Find the work W done when the point of application of F moves from P(- 3, 7) to Q(10, -1). (Assume distance measured in meters.)

Solution

F = 30(3i - 4j)/sqrt(3² + 4²) = 6(3i - 4j)

dr = (10-(-3))i + (-1 - 7)j = 13i - 8j

work, W = F.dr = 6(3i - 4j).(13i - 8j) = 6 x (39 + 32) = 426 J