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.
What is the maximum distance x1 that the block moves horizontally away from the track at x=0? (in meters)
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)
fall distance R = 2
Ke = (1/2) m v^2 = m g R
v = sqrt (2 g R) = 2 sqrt g =2 sqrt 10
normal force = m g = 2 g
friction force = mu m g = -2 g mu
mu = 1.1 x
so friction force = -2.2 g x
F = m a
m a = -2.2 g x - k x
d^2x/dt^2 = -2.2 g x = -22 x - 10 x = -32 x
dx/dt = Vinitial - 16 x^2
when is v = dx/dt = 0?
-16 x^2 = Vinitial = 2 sqrt (10)
that gives you x when v = 0
integrate again for t
To find the maximum distance the block moves horizontally away from the track, x1, we need to consider the forces acting on the block at that point. At the bottom of the track, the block has converted its potential energy to kinetic energy, so the sum of the gravitational force and the force from the compressed spring must provide enough negative work to bring the block to a stop. Let's break down the steps to find x1:
1. Find the gravitational potential energy at the top of the track:
The gravitational potential energy is given by U = m * g * h, where m is the mass of the block and h is the height at the top of the track. In this case, h = R, so U = m * g * R.
2. Find the force exerted by the compressed spring:
The force exerted by the compressed spring is given by F = k * x, where k is the spring constant and x is the compression of the spring. At the maximum distance x1, the spring is fully compressed, so F = k * x1.
3. Find the work done by the spring:
The work done by the spring is given by W = (1/2) * k * x1^2, where k is the spring constant and x1 is the maximum distance. This work is equal to the negative of the gravitational potential energy at the top of the track, so (1/2) * k * x1^2 = -m * g * R.
4. Solve for x1:
Setting the work equation equal to the gravitational potential energy equation, we have (1/2) * k * x1^2 = -m * g * R. Plugging in the given values, we have (1/2) * 10 * x1^2 = -1 * 10 * 2. Solving for x1, we find x1 = √4 = 2 meters.
Therefore, the maximum distance, x1, that the block moves horizontally away from the track is 2 meters.
To find the time it takes for the block to travel between x = 0 (relaxed spring) and x = x1 (block at the first stop), we need to calculate the time it takes for the block to reach x1 using the principles of motion.
1. Find the acceleration of the block:
The net force acting on the block in the horizontal direction is given by F_net = m * a, where m is the mass of the block and a is its acceleration. At x = 0, the force of the compressed spring is balanced by the force of static friction, so F_net = k * x1 = μ(0) * m * g. Plugging in the given values, we have 10 * 2 = α * 0 * 10 * 1. Solving for α, we find α = 1.1 m^(-1).
2. Find the coefficient of friction at x = x1:
The coefficient of friction at x = x1 is given by μ(x1) = α * x1. Plugging in the values we found earlier, μ(x1) = 1.1 * 2 = 2.2.
3. Find the acceleration of the block at x = x1:
The net force at x = x1 is given by F_net = m * a, where a is the acceleration at x = x1. The net force is provided by the force of kinetic friction acting in the opposite direction to the motion, so F_net = μ(x1) * m * g = -α * x1 * m * g.
4. Solve for the acceleration at x = x1:
Plugging in the values, we have -α * x1 * m * g = 2.2 * 2 * 1 * 10 = -44 m/s^2. Since the acceleration is negative, it means the block decelerates until it comes to a stop.
NOTE: Because the friction force is not uniform, the block won't move from x = 0 to x = x1 with constant acceleration. We can't use the simple equations of motion like s = ut + (1/2)at^2 and v = u + at directly. Another approach is needed.
5. Find the time t1 using calculus:
To find the time t1, we need to find the integral of the acceleration with respect to time from x = 0 to x = x1. This will give us the time needed for the block to travel from x = 0 to x = x1.
∫(dv/dx)dt = ∫(-α * x1 * g)dt = -α * x1 * g * ∫dt = -α * x1 * g * t + C,
where C is the constant of integration.
To evaluate the indefinite integral, we need to find the bounds of integration. At t = 0, x = 0, and at t = t1, x = x1. So, the definite integral is:
∫[0 to t1](-α * x1 * g)dt = -α * x1 * g * t1.
Setting this equal to x1 and solving for t1, we have:
-α * x1 * g * t1 = x1,
-α * g * t1 = 1,
t1 = -1 / (α * g).
Plugging in the given values, we have t1 = -1 / (1.1 * 10) = -1 / 11 = -0.0909 seconds.
Since time can't be negative in this case, the negative sign doesn't make physical sense. Therefore, the time t1 it takes for the block to travel between x = 0 (relaxed spring) and x = x1 (block at the first stop) is approximately 0.0909 seconds.