Gliding mass stopped by spring and friction

(12 points possible)

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= 7 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=
0.7 m
−1. Assume for simplicity that static and dynamic friction coefficients are the same, and use
g=10
m/s
2. (See figure)

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

x
1=
unanswered

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

t
1=
unanswered

(c) What will happen after the block reaches point x
1?

The block will stay put forever at x=x1.
The block will move back and reach a second stop exactly at x=0.
The block will move back and reach a second stop somewhere between x=0 and x=x1.
The block will move back and get catapulted up the circular track.

(a) To find the maximal distance x1 that the block moves horizontally away from the track, we need to analyze the forces acting on the block at that position. At the point x1, the block experiences friction and the spring force.

The force of friction can be determined as follows:

Friction force (F_friction) = coefficient of friction (μ) * normal force (N)
Normal force (N) = mass (m) * acceleration due to gravity (g)
μ(x1) = α * x1 (given)
N = m * g

At the point x1, the spring force (F_spring) is in equilibrium with the friction force. Thus,

F_spring = F_friction

The spring force can be given as:

F_spring = k * x1 (Hooke's Law)

Equating the two forces, we have:

k * x1 = μ(x1) * m * g
7 * x1 = α * x1 * m * g

Plugging in the given values:
α = 0.7 m^(-1)
m = 1 kg
g = 10 m/s^2

7 * x1 = 0.7 * x1 * 1 * 10

Simplifying the equation:

7 * x1 = 7 * x1

This implies that x1 can have any value as long as it satisfies the given conditions. Therefore, the maximal distance x1 that the block moves horizontally away from the track is unanswered.

(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 can use the concept of work-energy theorem. The work done by the net force acting on the block is equal to its change in kinetic energy.

At x=0 (relaxed spring), the velocity of the block is 0, and at x=x1, the velocity of the block is also 0 (block at first stop). Therefore, the change in kinetic energy is 0.

The work done by the forces acting on the block is equal to the negative of the work done by friction (since it acts in the opposite direction of the displacement).

Work done by friction:

Work_friction = -μ(x1) * m * g * x1

Work done by the spring force:

Work_spring = (1/2) * k * x1^2

Equating the work done by the forces to 0, we have:

-μ(x1) * m * g * x1 + (1/2) * k * x1^2 = 0

Substituting the given values:
α = 0.7 m^(-1)
m = 1 kg
g = 10 m/s^2
k = 7 N/m

-0.7 * 1 * 10 * x1 + (1/2) * 7 * x1^2 = 0

Simplifying the equation:

-7x1 + 3.5x1^2 = 0

We can solve this quadratic equation for x1:

3.5x1^2 - 7x1 = 0
x1(3.5x1 - 7) = 0

This gives two possible solutions:
x1 = 0 (rejected, as it corresponds to the relaxed spring position)
x1 = 2 m

Therefore, the time t1 it takes for the block to travel between x=0 and x=x1 is unanswered.

(c) After the block reaches the point x1, it will move back and reach a second stop somewhere between x=0 and x=x1.

To answer these questions, we need to analyze the forces acting on the block as it moves along the track and encounters the spring and friction.

Let's break down the problem and solve each part step by step:

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

To determine the maximal distance x1, we need to find the point where the block stops moving horizontally. This occurs when the force of friction equals the force from the spring.

The force of friction can be calculated using the friction coefficient μ(x) = αx, where α = 0.7 m^(-1) is given. Since the block is moving horizontally, the friction force is acting in the opposite direction of motion. Thus, the friction force can be represented as F_friction = -μ(x) * mg, where mg is the weight of the block.

The force from the spring can be determined using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position. In this case, the displacement is x1. Thus, the force from the spring can be represented as F_spring = k * x1.

At the point where the block stops moving horizontally, the forces of friction and the spring are equal:

-μ(x1) * mg = k * x1

Substituting the given values, we get:

-0.7 * x1 * 1 * 10 = 7 * x1

Simplifying the equation, we have:

-0.7 * x1 = 7

Solving for x1, we find:

x1 = 10 meters

So, the maximal distance x1 that the block moves horizontally away from the track at x=0 is 10 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)?

To find the time t1, we need to use the equation of motion for the block. The force acting on the block in the horizontal direction is the force of friction, given by F_friction = -μ(x) * mg.

Using Newton's second law, F = ma, where a is the acceleration, we can write:

-μ(x) * mg = m * a

Since acceleration is the second derivative of position with respect to time, a = d^2x/dt^2, we can rewrite the equation as:

-μ(x) * mg = m * (d^2x/dt^2)

Substituting the given values of μ(x) and mg, we get:

-0.7 * x * 1 * 10 = 1 * (d^2x/dt^2)

Simplifying the equation, we have:

-0.7 * x = d^2x/dt^2

To find time t1, we need to integrate the equation of motion. Since the block starts from rest at x = 0, the initial conditions for integration are x(0) = 0 and dx/dt(0) = 0.

Integrating the equation, we get:

-0.7 * x = dx/dt

Integrating once more, we obtain:

-0.7 * x1 = x

To find the time t1, we need to solve for t when x = x1:

-0.7 * x1 = x1

Simplifying the equation, we have:

-0.7 = 1

This equation does not have a valid solution. It means that the block never reaches the point x = x1, and it keeps moving indefinitely.

Therefore, the time t1 is undefined since the block does not reach x1.

(c) What will happen after the block reaches point x1?

The correct answer is:

The block will move back and reach a second stop exactly at x = 0.

After reaching point x1, the block will experience a force in the opposite direction due to the compressed spring. This force will cause the block to move back towards the circular track. The magnitude of the force from the spring is still given by F_spring = k * x1.

Once the block reaches point x = 0, the force from the spring will be at its maximum, pushing the block back towards the track. The block will stop again when the force from the spring equals the force of friction, which is now acting in the opposite direction.

The block will continue oscillating between x1 and x = 0, reaching a stop at both points. This will repeat indefinitely, and the block will not be catapulted up the circular track.

So, the correct answer is that the block will move back and reach a second stop exactly at x = 0.