3: A dragster accelerates from rest down a street of length d = 400.0 m. In the absence of any friction the dragster has a constant acceleration of a = 28.1 m/s² in the direction of motion, and its mass is m = 1095 kg. Assume the dragster is moving in the positive horizontal direction.

Let's work through an example to find the power output of a motor. An electric motor is used to run an elevator. The total mass of the elevator car and passenger is 1350 kg. The elevator moves upward a distance of 27.0 m in 15.88 s at a constant speed. What is the power output of the motor during this time? Let's first find the amount of work done. From the mass of the elevator, we can find the weight of the elevator. Because the elevator is moving at a constant speed, the upward force that is applied to the elevator by the motor must be equal to the downward weight. What is the force (in N) on the elevator by the motor? F= x Think about Newton's second law. Consider the elevator moving upward at a constant speed. What is the acceleration if the speed is constant? What is the net force? If we know the net force, and we know the weight downward, what must be the force upward? N Now, knowing the force, and the distance the elevator moves, we can find the work done by the motor. Remember…

The force component along the displacement varies with the magnitude of the displacement, as shown in the graph. Find the work done by the force in the interval from (a) 0 to 0.9 m, (b) 1.0 to 2.0 m, and (c) 2.0 to 3.8 m. (Note: In the last interval the force component is negative, so the work is negative.)

While riding on an elevator, a person has twoforces acting on him, a normal force and a gravity force. Suppose the person has mass m is on an elevator that 1 descends a distance d. If the total work on the person is Wiotal = -mgd, what is 4 the magnitude of the normal force? Is the elevator speeding up or slowing down? The knowns in this problem are m and g.