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.)
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 consider the forces acting on the block at different points and calculate the minimum energy required to complete the loop.
Let's break down the problem step by step:
1. The block is initially at rest and pressed against the spring, compressing it a distance "d". At this point, the block has the potential energy stored in the spring, given by:
U_spring = 0.5 * k * d^2 (1)
where k is the spring constant.
2. When the block is released, the spring begins to push it forward, and it moves from left to right on the track.
3. At point A, the block loses contact with the track, and the only forces acting on it are gravity and the normal force (vertical). There is no net force acting on the block in the horizontal direction.
4. Between points A and B, there is a force of kinetic friction acting on the block, opposing its motion. The kinetic friction force, F_friction, can be calculated using the coefficient of kinetic friction (μk) and the normal force (N):
F_friction = μk * N
The normal force N is equal to the weight of the block:
N = mg (2)
where m is the mass of the block, and g is the acceleration due to gravity.
Substituting equation (2) into the friction force equation gives:
F_friction = μk * mg (3)
5. The work done by the friction force over the distance AB is given by:
W_friction = F_friction * AB (4)
where AB is the length of the track section between points A and B.
Substituting equation (3) into equation (4), we have:
W_friction = μk * mg * AB (5)
This work done by friction represents the energy lost by the block as it moves from point A to point B.
6. At point B, the block has reached the end of the track, and now it must make it through the loop-the-loop.
7. At point C (top of the loop), the block is moving in a circular path with radius R. The centripetal force required to keep the block moving in a circle is provided by the normal force N and the gravitational force mg.
N - mg = mv^2 / R (6)
where v is the velocity of the block at point C.
8. At the top of the loop (point C), the normal force N is zero. For the block to barely make it through the loop, the normal force should approach zero, which means the block should just lose contact with the track at that point.
9. As the block reaches the top of the loop, all of its initial potential energy from the compressed spring is converted into kinetic energy and gravitational potential energy:
U_spring = KE + PE
KE = (1/2)mv^2
PE = mgR
Substituting the expressions for KE and PE into the equation, we have:
0.5 * k * d^2 = (1/2)mv^2 + mgR (7)
10. Solving for v^2 in equation (6):
v^2 = gR (8)
11. Substituting equation (8) into equation (7):
0.5 * k * d^2 = (1/2)m * gR + mgR
Simplifying:
k * d^2 = mgR(1 + 1)
k * d^2 = 2mgR (9)
12. Substituting the values provided in the problem:
k = 84.8 N/m
m = 0.5 kg
R = 1.6 m
μk = 0.3
AB = 2.2 m
g = 9.8 m/s^2
Solving equation (9) for d:
d^2 = (2mgR) / k
d = sqrt((2 * 0.5 * 9.8 * 1.6) / 84.8)
d ≈ 0.491 m (rounded to appropriate significant figures)
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 0.491 meters.