A massless spring of constant k =84.8 N/m is fixed on the left side of a level track. A block of mass m = 0.5 kg is pressed against the spring and compresses it a distance d, as in the figure below. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.6 m. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. Given that the coefficient of kinetic friction between the block and the track along AB is μk=0.3, and that the length of AB is 2.2 m, determine the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at point C. (Hint: The force exerted by the track on the block will be zero if the block barely makes it through the loop-the-loop.)

PE(spring) =PE + W(fr)

k•x²/2 =m•g•(2R) +μ•m•g•s.

x= sqrt{2(m•g• 2R +μ•m•g•s)/k} =
= sqrt{2m•g• (2R +μ•s)/k}=
=sqrt{2•0.5•9.8(3.2 + 0.3•2.2)/84.8}=
=0.668 m

To determine the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at point C, we need to consider the forces acting on the block and apply the principles of work and energy.

1. Start by identifying the forces acting on the block in this system:

- Weight force (mg): Pulling the block downwards.
- Normal force (N): The force exerted by the track perpendicular to the surface.
- Force of spring (Fs): The force exerted by the compressed spring pushing the block forwards.
- Friction force (fk): The force of kinetic friction acting along AB.

2. Next, consider the different sections of the track and loop:

- For the circular loop-the-loop, the only forces acting are the weight force and normal force. Since the track is frictionless, there is no friction force.
- For the section AB, in order for the block to just make it through the loop-the-loop, the net force acting on the block must be zero. This means that the force of the spring is equal and opposite to the kinetic friction force.

3. Apply the equation of motion to the section AB:

The force of kinetic friction (fk) can be calculated using the equation: fk = μk * N

The normal force (N) can be calculated as the perpendicular component of the weight force: N = mg * cos(θ)

The force of the spring (Fs) can be calculated using Hooke's Law: Fs = k * d

Since the net force is zero, we can write the equation:

0 = Fs - fk

Substituting the expressions for Fs and fk:

0 = k * d - μk * N

4. Convert the equation to eliminate the normal force:

Substitute the expression for N into the equation:

0 = k * d - μk * (mg * cos(θ))

Since mg * cos(θ) is the weight force in the horizontal direction (Fh), we can rewrite the equation as:

0 = k * d - μk * Fh

5. Substitute the values and solve for d:

Given:
k = 84.8 N/m,
m = 0.5 kg,
R = 1.6 m,
μk = 0.3,
AB = 2.2 m.

The horizontal component of the weight force (Fh) can be calculated using Fh = mg * sin(θ). Since the block is at rest initially, we can determine the angle θ using the radius R:

R = (AB + 2R) * sin(θ)
1.6 = (2.2 + 2 * 1.6) * sin(θ)
1.6 = 2.2 + 3.2 * sin(θ)

Solving for sin(θ), we find sin(θ) = (1.6 - 2.2) / 3.2 = -0.2.

Now, calculate Fh = mg * sin(θ):
Fh = (0.5 kg) * (9.8 m/s^2) * (-0.2) = -0.98 N

Substitute the known values into the equation:

0 = (84.8 N/m) * d - (0.3) * (-0.98 N)

Simplifying:

0 = 84.8d - 0.294

Rearranging to solve for d:

d = 0.294 / 84.8 ≈ 0.003461 m

Therefore, the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at point C is approximately 0.003461 m.