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!