A mass m = 73 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 17.9 m and finally a flat straight section at the same height as the center of the loop (17.9 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.)

What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track?

What height above the ground must the mass begin to make it around the loop-the-loop?

If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

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 (17.9 m off the ground)?

Now a spring with spring constant k = 15500 N/m is used on the final flat surface to stop the mass. How far does the spring compress?

To find the minimum speed required at the top of the loop, we can use the conservation of energy principle. At the top of the loop, the gravitational potential energy is converted into kinetic energy.

1. Minimum speed at the top of the loop:
To find the minimum speed, we can equate the gravitational potential energy at the starting height (above the top of the loop) to the sum of the gravitational potential energy at the top of the loop and the kinetic energy at that point.
The equation can be written as:
mgh = mgh + 0.5mv^2
Simplifying the equation:
gh = gh + 0.5v^2
0.5v^2 = gh
v^2 = 2gh
Taking the square root of both sides:
v = √(2gh)

Now we can plug in the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 17.9 m (height above the ground)

v = √(2 * 9.8 * 17.9)
v ≈ 26.26 m/s

So, the minimum speed the block must have at the top of the loop is approximately 26.26 m/s.

2. Height above the ground to begin making it around the loop-the-loop:
To determine this height, we can use the conservation of mechanical energy. We know that the minimum speed required at the top of the loop is the same as the speed at the highest point of the loop.
We can use the equation:

mgh1 + 0.5mv^2 = mgh2 + 0.5mv^2

Since the mass is released from a height h above the top of the loop, h1 will equal h + R, where R is the radius of the loop. The height at the top of the loop (h2) is R.
The equation can be simplified to:

gh1 + 0.5v^2 = gh2 + 0.5v^2
gh + gR + 0.5v^2 = gR + 0.5v^2
gh = gR
h = R

Therefore, the height above the ground from which the mass must be released to make it around the loop is equal to the radius of the loop, which is 17.9 m.

3. Speed at the bottom of the loop:
When the mass reaches the bottom of the loop, it will have converted its gravitational potential energy entirely into kinetic energy. This means that the speed at the bottom of the loop will be equal to the square root of 2gh, where h is the height of the center of the loop above the ground.
Using the given value of h = 17.9 m and substituting it into the formula:

v = √(2 * 9.8 * 17.9)
v ≈ 26.26 m/s

So, the speed at the bottom of the loop is approximately 26.26 m/s.

4. Speed at the final flat level:
Since the final flat level is at the same height as the center of the loop (17.9 m off the ground), the gravitational potential energy at the top of the loop is equal to the gravitational potential energy at the final flat level.
This means that the speed at the final flat level will also be the same as the speed at the top of the loop, which is approximately 26.26 m/s.

5. Distance the spring compresses:
To find the distance the spring compresses, we can use the conservation of mechanical energy again. The initial kinetic energy plus the initial potential energy will be equal to the potential energy stored in the compressed spring. The equation can be written as:
0.5mv^2 = 0.5kx^2

We already know the mass (m) and the spring constant (k), so we need to find x, the distance the spring compresses.
Rearranging the equation:

x^2 = (mv^2) / k
x = √((mv^2) / k)

Plugging in the given values:
m = 73 kg
v = 26.26 m/s
k = 15500 N/m

x = √((73 * 26.26^2) / 15500)
x ≈ 1.83 m

So, the spring compresses approximately 1.83 m.