A mass m = 85 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally a flat straight section at the same height as the center of the loop (19.2 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

iv) If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (19.2 m off the ground)?

To determine the speed of the mass at the final flat level, we can use the principle of conservation of energy. At the initial height (above the top of the loop), the mass possesses both gravitational potential energy and kinetic energy. At the final flat level, all the gravitational potential energy would have been converted to kinetic energy, assuming no energy losses due to friction or other factors.

Here are the steps to find the mass's speed at the final flat level:

1. Calculate the height of the initial position above the top of the loop:
- Given: Radius of the loop, R = 19.2 m
- The center of the loop is at the same height as the final flat level, which is 19.2 m off the ground.
- Therefore, the initial position is R + 19.2 m.

2. Calculate the gravitational potential energy at the initial position:
- Gravitational potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
- Gravitational acceleration (g) is approximately 9.8 m/s².
- Since the track is frictionless, there is no change in mechanical energy, so the gravitational potential energy at the initial position is equal to the kinetic energy at the final flat level.
- PE = m * g * h_initial

3. Use the conservation of mechanical energy principle:
- At the final flat level, the kinetic energy (KE) is equal to the gravitational potential energy at the initial position.
- KE = m * g * h_initial
- Since the mass has just enough speed to make it around the loop without leaving the track, the speed at the final flat level is the speed required to balance the gravitational force at the initial position.

4. Calculate the speed at the final flat level:
- Gravitational potential energy (PE) = m * g * h_initial = KE = (1/2) * m * v²
- Set the gravitational potential energy equal to the kinetic energy.
- Solve for the speed (v).

After calculating the above steps, you will find the speed (v) at the final flat level, which is the answer to the question.

To determine the speed of the mass at the final flat level, we can use the principle of conservation of mechanical energy. At the highest point of the loop (assuming it is released from there), the mass has gravitational potential energy equal to the initial kinetic energy. At the final flat level, all of the initial potential energy is converted into kinetic energy.

Let's denote the height from the top of the loop to the final flat level as h.

1) Find the gravitational potential energy at the highest point of the loop:
Potential Energy (PE) = mass (m) * gravity (g) * height (h)
PE = 85 kg * 9.8 m/s^2 * h

2) Find the speed (v) at the top of the loop using conservation of energy:
PE = KE
85 kg * 9.8 m/s^2 * h = (1/2) * mass (m) * v^2
Simplifying:
9.8 m/s^2 * h = (1/2) * v^2
v^2 = 19.6 m/s^2 * h
v = sqrt(19.6 m/s^2 * h)

3) Find the speed at the final flat level:
Since the height at the final flat level is the same as the center of the loop, the gravitational potential energy at this point is zero. Therefore, all the initial potential energy is converted into kinetic energy.
KE = (1/2) * mass (m) * v^2
KE = (1/2) * 85 kg * (sqrt(19.6 m/s^2 * h))^2
KE = (1/2) * 85 kg * 19.6 m/s^2 * h
KE = 833 kg m^2/s^2 * h

Since the kinetic energy at the final flat level is equal to the potential energy at the highest point of the loop, we have:
833 kg m^2/s^2 * h = 85 kg * 9.8 m/s^2 * h

Simplifying:
833 kg m^2/s^2 = 833.4 kg m^2/s^2 = 85 kg * 9.8 m/s^2
h = 833.4 kg m^2/s^2 / (85 kg * 9.8 m/s^2)
h ≈ 0.98 m

Finally, substitute the value of h into the expression for v:
v = sqrt(19.6 m/s^2 * h)
v = sqrt(19.6 m/s^2 * 0.98 m)
v ≈ 4.4 m/s

Therefore, the speed of the mass at the final flat level is approximately 4.4 m/s.