: 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.)

Question

: 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.)

Answer 1

To determine the minimum compression 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 at different points along the track.

Let's analyze the forces acting on the block at different points along the track and loop-the-loop system:

1. At Point A:
The only force acting on the block is the normal force (N1) exerted by the track on the block, which is equal to the weight of the block (mg) since the block is at rest.

N1 = mg

2. Between Points A and B:
The block experiences frictional force along this section. The force of friction (f1) is given by the coefficient of kinetic friction (μk) multiplied by the normal force (N1). The work done by the frictional force should be equal to the change in the potential energy of the spring.

f1 = μk * N1
Work done by frictional force (W1) = f1 * AB

3. At Point B:
The block will experience forces from both the spring and gravity. The compressing force exerted by the spring is given by Hooke's Law:

Fs = -k * d

At this point, the spring force (Fs) and the force of gravity (mg) are acting in the same direction.

Fs + mg = Net force

4. Between Point B and Point C (loop-the-loop):
In order for the block to successfully go through the loop-the-loop, the normal force acting on the block at Point C should be zero. This means that the net force at that point should be zero.

Net force = Centripetal force
Net force = Fs + mg + f2
Centripetal force = m * v^2 / R

Here, f2 is the frictional force acting on the block at Point C, equal to the coefficient of kinetic friction (μk) multiplied by the normal force at Point C.

f2 = μk * N2
N2 = mg + Fs

Thus,
Net force = Fs + mg + μk * (mg + Fs)
m * v^2 / R = Fs + mg + μk * (mg + Fs)

5. At Point C:
The normal force at this point should be zero for the block to just make it through the loop-the-loop.

Now, we can proceed to solve the problem. Substituting the values and equations mentioned above, we can solve for the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at Point C.