A swinging rope is used to transport 0.03kg of medicine. The rope has a length of 2.4m, so that the tension of the rope at the bottom point of the swing is 0.3929N. What is the tangential speed of the medicine at this bottom point? Answer in m/s.

Thanks.

To find the tangential speed of the medicine at the bottom point of the swing, we can use the concept of conservation of mechanical energy.

First, let's assume that at the bottom point of the swing, all the potential energy of the medicine has been converted into kinetic energy.

The potential energy of an object at height h is given by the formula:
Potential energy = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

In this case, the height can be considered as the vertical displacement of the medicine when it is at the bottom of the swing. Since the length of the rope is 2.4m, the vertical displacement will be half of that, or 1.2m.

Next, we can equate the potential energy to the kinetic energy at the bottom point:
Potential energy = Kinetic energy

m * g * h = (1/2) * m * v^2, where v is the tangential speed we are trying to find.

Since the mass (m) cancels out on both sides of the equation, we are left with:

g * h = (1/2) * v^2

Now we can substitute the given values:

(9.8 m/s^2) * (1.2 m) = (1/2) * v^2

11.76 = 0.5 * v^2

Dividing both sides by 0.5:

v^2 = 11.76 / 0.5

v^2 = 23.52

Taking the square root of both sides:

v ≈ √23.52

v ≈ 4.85 m/s

Therefore, the tangential speed of the medicine at the bottom point of the swing is approximately 4.85 m/s.