Question

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.

Answer 1

Answer:Sorry i cannot draw a diagram but ill try to explain it my best.

Step-by-step explanation:

To find the final velocity of the paper, we can use the principle of conservation of energy. At the start, the paper has potential energy due to its position on the refrigerator door, which is gradually converted into kinetic energy as it slides down due to the force of friction. We can write:

Initial potential energy = Final kinetic energy

mgh = (1/2)mv^2

where m is the mass of the paper, g is the acceleration due to gravity, h is the initial height of the paper on the door, and v is the final velocity of the paper.

We can rearrange this equation to solve for v:

v = sqrt(2gh)

where sqrt means "square root".

To calculate the height h, we can use the fact that the paper slides down a distance of 1.33 m. We know that the force of friction opposing the motion is:

Ffriction = μk * Fnormal

where Fnormal is the normal force, which is equal in magnitude to the force applied by the magnet.

Fnormal = 1.071 N

Ffriction = μk * Fnormal = 0.213 * 1.071 N = 0.228 N

The force of gravity pulling the paper down is:

Fgravity = m * g = 0.07 kg * 9.81 m/s^2 = 0.6867 N

The net force on the paper is:

Fnet = Fapplied - Ffriction - Fgravity

where Fapplied is the force applied by the magnet.

Fnet = 1.071 N - 0.228 N - 0.6867 N = 0.1563 N

This net force causes the paper to accelerate down the door. The acceleration is:

a = Fnet / m = 0.1563 N / 0.07 kg = 2.233 m/s^2

Using the kinematic equation:

v^2 = v0^2 + 2ad

where v0 is the initial velocity (zero in this case), d is the distance traveled (1.33 m), and a is the acceleration, we can solve for v:

v = sqrt(2ad) = sqrt(2 * 2.233 m/s^2 * 1.33 m) = 2.79 m/s

Therefore, the paper is moving at a speed of 2.79 m/s (rounded to two significant figures) when it reaches the bottom of the door.