A massless spring of constant k = 91.0 N/m is fixed on the left side of a level track. A block of mass m = 0.50 kg is pressed against the spring and compresses it a distance of d. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.5 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 = 0.24, and that the length of AB is 2.5 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 of the track on the block will be zero if the block barely makes it through the-loop-the-loop.]
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To determine the minimum compression d of the spring that enables the block to just make it through the loop-the-loop, we need to analyze the forces acting on the block.
1. First, let's consider the block at point A, where it is in contact with the spring. The spring exerts a force on the block given by Hooke's Law: F_spring = -k * d, where k is the spring constant and d is the compression distance.
2. The block's weight (mg) acts vertically downward and can be divided into two components: one parallel to the track (mg * sinθ) and one perpendicular to the track (mg * cosθ), where θ is the angle formed by the track at point A.
3. The normal force (N) exerted by the track on the block at point A is equal in magnitude but opposite in direction to the perpendicular component of the weight (mg * cosθ).
4. As the block moves along the track from point A to point B, there is a frictional force opposing its motion given by F_friction = μk * N, where μk is the coefficient of kinetic friction.
5. At point B, just before the loop-the-loop, the block loses contact with the track momentarily, and the normal force becomes zero.
6. At the top of the loop-the-loop (point C), with the block moving in a vertical circular path, the net force acting on the block must be directed toward the center of the loop-the-loop to keep it in circular motion.
Now, let's apply the conditions for the block to barely make it through the loop-the-loop:
a) At point B, the block just loses contact with the track, meaning that the net vertical force is zero. Therefore, the sum of the vertical components of the weight and the normal force must add up to zero:
mg * cosθ - N = 0
b) At point C, the block is moving in a vertical circular path, so the net force points toward the center of the loop-the-loop. This means the net force is the difference between the vertical components of the weight and the normal force:
N - mg * cosθ = mv^2 / R
Since we are looking for the minimum compression d of the spring, we want to consider the most favorable conditions for the block to make it through the loop-the-loop. This occurs when the block has the minimum speed at the top of the loop-the-loop, which happens when it barely makes it.
c) The minimum speed for the block to barely make it through the loop-the-loop can be found using conservation of mechanical energy. Initially, at point A, the block has potential energy due to the compression of the spring:
PE_initial = (1/2) * k * d^2
At the top of the loop-the-loop (point C), the block has kinetic energy and gravitational potential energy:
KE_final + PE_final = (1/2) * m * v^2 + m * g * 2R
Since the block barely makes it through the loop, its net vertical force at point B is zero, and thus it is momentarily weightless. This means that its gravitational potential energy at point C is equal to zero:
KE_final + 0 = (1/2) * m * v^2
Now we can solve the equations.
Combining a) and b), we have:
mg * cosθ - mg * cosθ = mv^2 / R
0 = mv^2 / R
Dividing by m on both sides gives us:
0 = v^2 / R
And since the net vertical force at point B is zero, combining a) and b) we have:
mg * cosθ = mg * cosθ
N = mg * cosθ
Since N = mg * cosθ and N - mg * cosθ = mv^2 / R:
mv^2 / R = mg * cosθ
Cancelling out the mass:
v^2 / R = g * cosθ
Therefore, the minimum speed v at the top of the loop-the-loop is given by:
v^2 = g * R * cosθ
Recall that the minimum speed corresponds to when the block barely makes it through the loop. At this point, the net vertical force applied by the track (normal force) is zero, and thus the net horizontal force is also zero.
d) The net horizontal force at point B is given by:
F_net_horizontal = F_spring + F_friction
For the block to barely make it through the loop, the net horizontal force at point B must be zero:
-F_k - μk * N = 0
Plugging in the values for the forces, we have:
-k * d - μk * mg * cosθ = 0
Rearranging the equation, we can solve for the compression distance d:
d = -μk * mg * cosθ / k
Substituting the expression for N from a):
d = -μk * mg * cosθ / k = -μk * (mg * cosθ) / k = -μk * N / k
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 given by:
d = -μk * N / k
Note: The negative sign in the equation arises from the fact that the compression distance of the spring is considered negative when the block is compressed.