[l IRV page 13 of 490 d Example 6-1: In the situation pictured above, assume the mass of the crate,m, is 20 kg and the coefficient of kinetic friction between the crate and the floor is 0.4. IfF = 100 Nandd = 6 m, a) How much work is done by F? b) How much work is done by the normal force? ) How much work is done by gravity? d) How much work is done by the force of friction? ) What is the total work done on the crate? Solution: a) Because F is parallel to d, the work done by F is simply Fd = (100 N)(6 m) = 600 J. b) The normal force is perpendicular to the floor, and tod. Since the angle between F and d is 8 = 90, and cos 90" = 0, the work done by F, is zero. ) The gravitational force is also perpendicular to the floor, and tod. Because the angle between F,, and d is@ = 90, and cos 90 = 0, the work done by Fgm is zero, too. d) First, since F, =mg = (20 kg)(10 N/kg) = 200 N, we have F = nF, = (0.4)(200 N) = 80 N. However, the direction of the vector F, is opposite to the direction of d, so the angle between F; and d is 8 = 180. Because cos 180" = -1, the work done by the friction force is (80 N) (6 m)(-1) = -480 J. e) To find the total work done on the crate, we just add up the work done by each of the forces that acts on the crate, In this case, then, we'd have Weom = W + W, + W v F total by Fy, by F, + Wby? = (600J) + (0J)+ (0 J)+ ( 480J) = 120J e Here are a couple of things to notice aboutExample 6-1: 1) Although work depends on two vectors for its definition (namely,F and d), work itself is not a vector. Work is a scalar. W may be positive, negative, or zero, but work has no direction. 2) In this example, there were four forces acting on the crate: the pushing forceF, gravity, the normal force, and friction. Each force does its own amount of work, which is why each part had to specify for which force we wanted the work. Only in the last part, where the total work is desired, can we omit the specific force we're locking at (because we're considering them all). Example 6-2: In the situation described in Example 6-1, what is the net force on the crate? How much work is done byF_? net Solution: The normal force cancels out the gravitational force, so the net force on the crate is justF + F; = (100 N) + (80 N) = +20 N, where the + indicates thatF,, points to the right. Now, since E,, is parallel to d, the work done by F_, is just the product, F d = (20 N)(6 m) = 120 J. Notice that this is the same as the total amount of work done on the crate, as we figured out in part (e) ofExample 6-1. This wasn't a coincidence. The total work done (found by adding up the values of the work done by each force separately) is always equal to the work done by the net force. Remember that work is a scalar and it can be positive, zero, or negative. Now here's how to knowwhen W will be positive, zero, or negative. Because W = Fd cos 0, and F and d are magnitudes (which means they're positive), the sign of W depends entirelv on the sign of cos 0. 1 O <