A 10 kg textbook lies on the floor of an elevator moving at 0.25 m/s when the elevator begins to slow down and eventually stops. If the normal force between the book and the floor during this event is 103 N, how long does it take the elevator to come to a stop? Which direction was the elevator moving?

The elevator was moving upwards. By applying Newton's laws of motion, we determined that the net force acting on the textbook was upward, which means the elevator was slowing down from an upward motion. It takes 0.5 seconds for the elevator to come to a stop.

Explanation:

This problem is based on the topic of Newton's laws of motion in physics. After considering the downward force of gravity and the upwards force of the elevator floor, we can calculate the net force acting on the book. The total gravity acting on the book is 10 kg * 9.8 m/s^2 = 98 N. Based on the given Normal force (103 N), the net force acting on the book is then 103 N - 98 N = 5 N upwards. This means the elevator was moving upwards as well, since it's slowing down to a stop.

To calculate the time taken for the elevator to stop, we use the formula F = m*a, where F is the force, m is the mass, and a is the acceleration. Solving for a, we get a = F/m = 5 N / 10 kg = 0.5 m/s^2. To get the time t, we divide the original speed of the elevator by the deceleration, so t = 0.25 m/s / 0.5 m/s^2 = 0.5 seconds. Therefore, it takes 0.5 seconds for the elevator to come to a stop.

Learn more about Newton's laws of motion here:

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