#$&* course PHY 231 11/09 12:03 pm 025. `query 25*********************************************
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Given Solution: `aSTUDENT SOLUTION AND INSTRUCTOR COMMENT: The magnitude of kinetic friction force is fk = mu-sub k * N. First we add the 1.9 to 9.8 and get 11.7 as the acceleration and times that by the 28 kg and get 327.6 as the force so plugging in we get fk = 0.32 * 327.6 = 104.8 N. ** Good. The net force Fnet on the box is Fnet = m a = 1.90 m/s^2 * 28 kg. The net force is equal to the sum of the forces acting on the box, which include the weight mg acting downward and the force of the floor on the box acting upward. So we have Fnet = Ffloor - m g = m a. Thus Ffloor = m g + m a = 28 kg * 9.8 m/s^2 + 28 kg * 1.90 m/s^2 = 28 kg * 11.7 m/s^2 = 330 N, approx. Being pushed at constant speed the frictional force is f = `mu * N, where N is the normal force between the box and the floor. So we have f = .32 * 330 Newtons = 100 N, approx. ** STUDENT QUESTION: I don't understand why the net force is the weight of the box + the upward force of the elevator. Since the weight is directed downwards, and the elevator is going upwards, shouldn't it be the force of the elevator - weight? ???? INSTRUCTOR RESPONSE There are two forces acting on the box in the vertical direction, the weight (acting downward) and the normal force exerted by floor of the the elevator on the box (which acts upward). Using Ffloor for the normal force, the net vertical force is therefore net vertical force = Ffloor - weight. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: