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= 10 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= 1.1 m−1. Assume for simplicity that static and kinetic friction coefficients are the same, and use g=10 m/s2.

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

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)

To solve this problem, we need to consider two phases of motion: the motion while the block is on the circular track, and the motion while it is moving horizontally after hitting the spring.

Let's start by analyzing the motion on the circular track:

1. Motion on the Circular Track:
Since there is no friction on the circular track, the only force acting on the block is its weight, which is directed downward. At the top of the track, the normal force is equal to the weight, N = mg = 1 kg * 10 m/s^2 = 10 N.

Using the concept of centripetal force, we can write the equation:
m * (V^2 / R) = N
where V is the velocity of the block.

Solving for V, we find:
V = sqrt(N * R / m) = sqrt(10 N * 2 m / 1 kg) = sqrt(20) ≈ 4.47 m/s

2. Motion after hitting the spring:
The block will compress the spring due to the horizontal motion. The spring will provide a force in the opposite direction to the movement, which will eventually stop the block.

The net force acting on the block is given by:
F_net = F_spring - F_friction = -kx - f
where F_spring is the spring force, F_friction is the friction force, k is the spring constant, x is the compression of the spring, and f is the friction force.

The friction force is given by:
f = μmg = αxmg

Substituting this into the net force equation, we get:
F_net = -kx - αmgx

Using Newton's second law, F_net = ma, where a is the acceleration, we can write:
ma = -kx - αmgx

Simplifying, we find the differential equation that describes the motion of the block:
m * d^2x/dt^2 + (k + αmg) * x = 0

Now, we can solve this differential equation to find the maximum distance and the time it takes to reach that distance.

a. Maximum Distance x1:
To find the maximum distance x1, we can use the conservation of mechanical energy. At the top of the track, the potential energy is entirely converted into kinetic energy. Therefore, we have:

mgh = (1/2)mv^2

Substituting the known values, we get:
1 kg * 10 m/s^2 * 2 m = (1/2) * 1 kg * v^2
20 J = (1/2) v^2
v = sqrt(40) ≈ 6.32 m/s

This is the initial velocity of the block after leaving the track.

Next, we need to find the maximum compression of the spring, xmax. We can use the conservation of mechanical energy again:

(1/2)mv^2 = (1/2)kx^2

Substituting the values, we have:
(1/2) * 1 kg * (sqrt(40))^2 = (1/2) * 10 N/m * (xmax)^2
20 J = 5 N/m * xmax^2
xmax = sqrt(4) ≈ 2 m

This is the maximum compression of the spring, which indicates the maximum distance x1 that the block moves horizontally away from the track.

Therefore, the maximum distance x1 is 2 m.

b. Time t1:
To find the time t1, we can use the equation of motion for the block on the horizontal surface:

ma = -kx - αmgx

Rearranging this equation, we have:
m * d^2x/dt^2 = -kx - αmgx

This is a second-order linear differential equation, and its solution will give us the position x as a function of time t.

Solving this differential equation, we get:
x(t) = Acos(ωt) + Bsin(ωt)
where A and B are constants determined by the initial conditions, and ω is the angular frequency given by:
ω = sqrt(k + αmg / m)

The initial conditions are:
x(0) = 0 (the block starts from rest at x = 0)
dx/dt(0) = v (the initial velocity of the block after leaving the track)

Applying these initial conditions, we find:
x(t) = xmax * cos(ωt)
where xmax is the maximum compression of the spring.

To find the time t1 for the block to travel between x = 0 and x = x1, we need to find the value of t when x = x1. Substituting this into the equation, we have:

x1 = xmax * cos(ωt1)

Simplifying this equation, we find:
t1 = acos(x1 / xmax) / ω

Substituting the known values, we get:
t1 = acos(2 / 2) / (sqrt(10 N/m + 1.1 m^-1 * 1 kg * 10 m/s^2 / 1 kg))
t1 = acos(1) / (sqrt(10 + 11))
t1 = 0 / (sqrt(21))
t1 = 0

Therefore, the time t1 is 0 seconds.

To summarize:
a. The maximum distance x1 that the block moves horizontally away from the track is 2 meters.
b. The time t1 it takes for the block to travel between x = 0 (relaxed spring) and x = x1 (block at the first stop) is 0 seconds.