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A piece of paper is placed on the door of a refrigerator with a magnet. The magnet is not very strong and can only supply a force of 1.071 N onto the paper. This direction of this force is directly towards the refrigerator door. This force is not strong enough to hold the paper stationary, and the paper starts to slide down the door. The paper is very thick and has a mass of 70.0 g. And there is a coefficient of friction between the paper and the door of μk = 0.213. (You can assume the magnet has zero mass). The paper slides down the door 1.33 m before reaching the bottom. How fast is the paper moving when it reaches the bottom? Use work-energy to solve this question. Your written work should include a properly drawn and labeled diagram, and properly labeled variables, and all values and variables must be properly derived as done in class.

User Niel Ryan
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Answer/Explanation: We can start by finding the force required to keep the paper from sliding down the door:

Weight of paper = mass x gravity = 0.07 kg x 9.81 m/s^2 = 0.6867 N

Force of friction = coefficient of friction x normal force

where the normal force is equal to the weight of the paper, since it's resting on the door.

Hence, force of friction = 0.213 x 0.6867 N = 0.1464 N

To keep the paper stationary on the door, we need a force that is equal to or greater than the sum of the weight of the paper and the force of friction:

Force required = weight of paper + force of friction = 0.6867 N + 0.1464 N = 0.8331 N

Since the magnet only supplies a force of 1.071 N, which is greater than the force required, the paper will stick to the door.

Now, as the paper slides down the door, the gravitational force acting on the paper does work on it, increasing its kinetic energy.

We can use the work-energy theorem:

Net work done on the paper = change in kinetic energy

The work done on the paper is the force of gravity times the distance slid:

Work done = force of gravity x distance = weight of paper x distance = 0.6867 N x 1.33 m = 0.913 Nm

The change in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy. Since the paper starts from rest, the initial kinetic energy is zero. So we have:

Net work done = (1/2)mv^2 - 0

Where m is the mass of the paper, and v is the final velocity of the paper.

We can equate the work done to the change in kinetic energy and solve for the final velocity:

0.913 Nm = (1/2) x 0.07 kg x v^2

v = sqrt[(2 x 0.913 Nm) / 0.07 kg] = 3.86 m/s

Therefore, the final velocity of the paper when it reaches the bottom of the door is 3.86 m/s.

Hope this helps, and have a good day!