Work Done by Constant Forces and Contact Forces The simplest work to evaluate is that done by a force that is constant in magnitude and direction. In this case, we can factor out the force; the remaining integral is just the total displacement, which only depends on the end points A and B, but not on the path between them: WAB = F. d = F . (FB -FA) = F|F8 -FA COse (constant force). We can also see this by writing this out in Cartesian coordinates and using the fact that the components of the force are constant: WAB [ F.dr = path AB path AB 1 (F2dr + F,dy + F.dz) = F. [ dt + Ry dy + F. [ª dz = F (x3 -xA) + Fy (VB - VA) + F2(ZB - ZA) = F . (FB - FA). Figure (a) below shows a person exerting a constant force F along the handle of a lawn mower, which makes an angle 0 with the horizontal. The horizontal displacement of the lawn mower, over which the force acts, is d. The work done on the lawn mower isW = F . d = Fd cos 8, which the figure also illustrates as the horizontal component of the force times the magnitude of the displacement. W = Fd cos 0 F cos 0 d e F (a) F F 0 = 90° cos 0 = 0 a d= 0 (h) (c)
Work done by a constant force. (a) A person pushes a lawn mower with a constant force. The component of the force parallel to the displacement is the work done, as shown in the equation in the figure. (b) A person holds a briefcase. No work is done because the displacement is zero. (c) The person in (b) walks horizontally while holding the briefcase. No work is done because cos 0 is zero. Figure (b) above shows a person holding a briefcase. The person must exert an upward force, equal in magnitude to the weight of the briefcase, but this force does no work, because the displacement over which it acts is zero. In figure (c), where the person in (b) is walking horizontally with constant speed, the work done by the person on the briefcase is still zero, but now because the angle between the force exerted and the displacement is 90" (F perpendicular to d' ) and cos 90° = 0. Calculating the Work You Do to Push a Lawn Mower How much work is done on the lawn mower by the person in figure (a) if he exerts a constant force of 75.0 N at an angle 35° below the horizontal and pushes the mower 25.0 m on level ground? Strategy We can solve this problem by substituting the given values into the definition of work done on an object by a constant force, stated in the equation W = Fd cos 8. The force, angle, and displacement are given, so that only the