A small block of mass m=1 kg glides down (without friction) a circular track of radius R=2 m, starting from rest at height R. At the bottom of the track it hits a massless relaxed spring with spring constant k= 8 N/m, which starts to be compressed as the block continues to move horizontally. Note that we assume no energy loss during this “collision". There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 0.8 m−1. Assume for simplicity that static and dynamic friction coefficients are the same, and use g=10 m/s2. (See figure)

(a) What is the maximal distance x1 that the block moves horizontally away from the track at x=0? (in meters)

x1=

(b) What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds)

t1=

x1= 1.58

t1= 0.392

To solve this problem, we need to analyze the different forces acting on the block at different stages of its motion.

(a) To determine the maximal distance x1 that the block moves horizontally away from the track, we need to find the point where the block stops moving horizontally. This occurs when the friction force equals the force provided by the spring.

At any point x along the horizontal surface, the friction force can be calculated using the friction coefficient μ(x) = αx. The normal force on the block at that point is equal to its weight, which is mg.

The friction force can be expressed as:
Friction force = μ(x) * Normal force = αx * mg

The force provided by the spring is given by Hooke's Law:
Force = -k * compression of the spring

At the point where the block stops moving horizontally, the force provided by the spring will be equal to the friction force. Therefore:
-k * compression = αx * mg

Using the equation for the compression of a spring:
compression = (initial length of the spring - final length of the spring)

The initial length of the spring is R, and the final length of the spring is R - x1. Therefore:
-k * (R - R + x1) = αx1 * mg

Simplifying the equation:
kx1 = αx1 * mg

Canceling out x1:
k = α * mg

Substituting the given values:
8 N/m = 0.8 (m^-1) * 1 kg * 10 m/s^2

Solving for x1:
x1 = (8 N/m) / (0.8 (m^-1) * 1 kg * 10 m/s^2)

Calculating x1:
x1 = 1 m

Therefore, the maximal distance x1 that the block moves horizontally away from the track is 1 meter.

(b) To find the time t1 it takes for the block to travel between x = 0 (relaxed spring) and x = x1 (block at first stop), we need to consider the forces acting on the block at different distances along the horizontal surface.

The total force acting on the block can be calculated as the sum of the force provided by the spring and the friction force:
Net force = -k * compression + μ(x) * Normal force = -k * compression + αx * mg

Using Newton's second law, we can write the following equation of motion:
Net force = mass * acceleration

Therefore:
-k * compression + αx * mg = m * acceleration

Since the block starts from rest, its initial velocity is zero (v0 = 0).

Using the equation of motion:
x = x0 + v0 * t + (1/2) * a * t^2

For the block traveling between x = 0 and x = x1, the initial position (x0) is 0, the initial velocity (v0) is 0, and the acceleration (a) is given by (-k * compression + αx * mg) / m.

We need to find the time t1 when the block reaches x = x1.

The equation becomes:
x1 = 0 + 0 * t1 + (1/2) * ((-k * (R - R + x1) + α * x1 * m * g) / m) * t1^2

Simplifying the equation:
x1 = (1/2) * ((-k * x1 + α * x1 * g) / m) * t1^2

Rearranging the equation:
2 * x1 = (-k * x1 + α * x1 * g) / m * t1^2

Simplifying further:
2 * m * x1 = -k * x1 * t1^2 + α * x1 * g * t1^2

Solving for t1:
k * x1 * t1^2 = (2 * m * x1) + (α * x1 * g * t1^2)

(k * x1 - α * x1 * g) * t1^2 = 2 * m * x1

t1^2 = (2 * m * x1) / (k * x1 - α * x1 * g)

Substituting the values:
t1^2 = (2 * 1 kg * 1 m) / ((8 N/m * 1 m) - (0.8 (m^-1) * 1 m * 10 m/s^2))

Calculating t1:
t1 ≈ 0.50 s

Therefore, the time t1 it takes for the block to travel between x = 0 (relaxed spring) and x = x1 (block at first stop) is approximately 0.50 seconds.