A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 m by a trained dog that exerts a horizontal force with magnitude 36.0 N. Use the work–energy theorem to find the final speed of the 12-pack if (a) there is no friction between the 12-pack and the floor, and (b) the coefficient of kinetic friction between the 12-pack and the floor is 0.30.
Solution 25E Introduction In this problem we will first calculate the net work done on the packet by the force applied on the packet. The net work done will be converted to the kinetic energy. By equating the work done with kinetic energy we will calculate the final velocity. Step 1 If F is the applied force and d is the displacement, then the work done by the force is given by W = Fdcos When there is no friction, the net work done will be equal to the work done by the force applied by the dogs. In this case we have F = 36.0 N 1 d = 1.20 m And angle = 0° Hence the work done is W = F dcos = (36.0 N)(1.20 m)cos(0°) = 43.2 J 1 1 Now suppose v i1 the velocity of the pack, then the kinetic energy is 1 2 1 2 Ek1 = 2v 1 = (2.30 kg)v 1 = 43.2 J 2(43.2 J) v =1 (4.30 kg)4.48 m/s So, if there is no friction, the velocity of the pack will be 4.48 m/s.