4. 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/s^2
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)
I was able to figure out part a, but I'm stuck on part b. Any help is appreciated
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 consider the forces acting on the block and use the laws of motion.
Let's analyze the forces acting on the block at each position:
1. At x = 0 (relaxed spring):
- The only force acting on the block is the force of friction f1, which is given by f1 = μ(x) * N.
- The normal force N is equal to the weight of the block, since there is no vertical acceleration.
- The weight W = m * g, where g is the acceleration due to gravity.
2. At x = x1 (block at the first stop):
- In addition to the force of friction f2, the block also experiences a spring force Fs due to compression of the spring.
Now, let's break down the steps to find the time it takes for the block to travel between these two positions:
Step 1: Find the maximum friction force f1(max) at x = 0:
- The coefficient of friction μ(x) = αx, where α = 1.1 m^(-1).
- At x = 0, we have μ(x) = α * 0 = 0.
- Therefore, the friction force at x = 0 is f1(max) = μ(x) * N = 0.
Step 2: Find the compression of the spring Δx at x = x1:
- The spring force Fs is given by Fs = k * Δx, where k is the spring constant.
- The work done by the spring force is equal to the work done against friction: Fs * Δx = f1(max) * x1.
- Substituting the known values, we get k * Δx^2 = 0. (since f1(max) = 0 at x = 0)
- Therefore, we have Δx = 0.
Step 3: Calculate the acceleration at x = x1:
- At x = x1, the net force acting on the block is given by Fnet = f2 - f1(max), where f2 is the friction force at x = x1.
- We can express f2 in terms of x1 using μ(x) = αx and f1(max) = 0 as f2 = μ(x) * N = α * x1 * N.
- Therefore, Fnet = α * x1 * N.
- The acceleration a at x = x1 is given by Fnet = m * a.
- Substituting the known values, we have α * x1 * N = m * a.
Step 4: Solve for x1:
- We can find N by equating the weight W and the normal force N, where W = m * g.
- The equation becomes m * g = N.
- Substituting N into the equation α * x1 * N = m * a, we get α * x1 * m * g = m * a.
- Simplifying, we have α * x1 * g = a.
- Using the kinematic equation v^2 = u^2 + 2a * s, where u is the initial velocity (0) and v is the final velocity (0),
and s is the distance x1, we have 0 = 0^2 + 2 * α * x1 * g * x1.
- Solving for x1, we get x1^2 = 0, which implies x1 = 0.
Since x1 = 0, the block does not move horizontally away from the track at x = 0. Therefore, the time t1 is also 0 seconds.
So, the answer to part b) is t1 = 0 seconds.