MASS PUSHED BY A SPRING
A block of mass m=2 kg on a horizontal surface is connected to a spring connected to a wall (see figure). The spring has a spring constant k= 16 N/m. The static friction coefficient between the block and the surface is μs= 0.7 , and the kinetic friction coefficient is μk= 0.5 . Use g=10 m/s2 for the gravitational acceleration.
(a) The spring is initially uncompressed and the block is at position x=0. What is the minimum distance x1 we have to compress the spring for the block to start moving when released? (in meters)
(b) Find the distance |x2−x1| between the point of release x1 found in (a), and the point x2 where the block will come to a stop again. (in meters)
(c) What time t12 does it take the block to come to a rest after the release? (i.e., the time of travel between points x1 and x2; in seconds)
(d) What will happen after the block has come to a rest at point x2?
The block will move back towards x1, and it will oscillate with constant frequency and exponentially decreasing amplitude.The block will move back towards x1, and it will oscillate while decreasing both frequency and amplitude.
The block will start moving back towards x1, and it will come to a final halt before reaching it.
The block will stay at its resting position x2.
The answer depends on whether x2>0 or x2<0
To solve this problem, we can use the principles of force and motion. Let's break down each question step-by-step:
(a) To find the minimum distance x1 we need to compress the spring for the block to start moving when released, we need to consider the forces acting on the block. When the block is at rest, the static friction between the block and the surface counterbalances the force exerted by the compressed spring. Using Newton's second law:
μs * m * g = k * x1
Where:
μs = static friction coefficient = 0.7
m = mass of the block = 2 kg
g = gravitational acceleration = 10 m/s^2
k = spring constant = 16 N/m
x1 = distance the spring is compressed
Rearranging the equation, we can solve for x1:
x1 = (μs * m * g) / k
= (0.7 * 2 * 10) / 16
≈ 0.875 meters
Therefore, the minimum distance x1 we need to compress the spring is approximately 0.875 meters.
(b) To find the distance |x2−x1| between the point of release x1 and the point x2 where the block will come to a stop again, we need to consider the forces acting on the block when it is moving. At point x1, the only force acting on the block is the force from the compressed spring, given by Hooke's Law:
Fs = -k * x1
At point x2, the block comes to a stop because the force of kinetic friction counterbalances the force from the spring. Using Newton's second law:
μk * m * g = k * x2
Where:
μk = kinetic friction coefficient = 0.5
m = mass of the block = 2 kg
g = gravitational acceleration = 10 m/s^2
x2 = point where the block comes to a stop
Rearranging the equation, we can solve for x2:
x2 = (μk * m * g) / k
= (0.5 * 2 * 10) / 16
≈ 0.625 meters
Therefore, the distance |x2−x1| between x1 and x2 is approximately 0.625 meters.
(c) To find the time t12 it takes for the block to come to a rest after the release, we can use the equation of motion:
x = x1 + v0*t + (1/2) * a * t^2
At point x2, the block has come to a stop, so its velocity v2 is 0. The acceleration a can be expressed in terms of the forces acting on the block:
Fs - μk * m * g = m * a
Substituting the force of the spring and rearranging the equation, we get:
-k * x2 - μk * m * g = m * a
Using the relation between force and acceleration:
a = (-k * x2 - μk * m * g) / m
Substituting back into the equation of motion and solving for t:
0 = x1 + v0 * t + (1/2) * [(-k * x2 - μk * m * g) / m] * t^2
Since v0 is 0, and rearranging the equation, we obtain a quadratic equation in terms of t:
(1/2) * [(-k * x2 - μk * m * g) / m] * t^2 = -x1
Simplifying:
t^2 = (-2 * x1 * m) / (-k * x2 - μk * m * g)
t = sqrt((2 * x1 * m) / (k * x2 + μk * m * g))
Substituting the given values:
t = sqrt((2 * 0.875 * 2) / (16 * 0.625 + 0.5 * 2 * 10))
≈ 0.52 seconds
Therefore, it takes approximately 0.52 seconds for the block to come to a rest after the release.
(d) After the block has come to a rest at point x2, it will move back towards x1 and will oscillate with constant frequency and exponentially decreasing amplitude.
To solve this problem, we need to analyze the forces acting on the block and use Newton's laws of motion.
(a) To find the minimum distance x1 to compress the spring for the block to start moving, we need to consider the forces involved. When the block is at rest, the static friction force between the block and the surface opposes the applied force. The maximum static friction force is given by μs * N, where N is the normal force acting on the block.
The normal force N is equal to the weight of the block, which is m * g, where g is the acceleration due to gravity. Therefore, the maximum static friction force is μs * m * g.
When the block starts moving, the static friction force changes to kinetic friction force, which is given by μk * N.
The force provided by the spring is given by Hooke's Law, which states that the force is proportional to the displacement. So, in this case, the force is given by F = -k * x, where k is the spring constant and x is the displacement.
At the point where the block starts moving, the force provided by the spring should be equal to the maximum static friction force. So, we can set up the equation: -k * x1 = μs * m * g.
Plugging in the given values, we have -16 * x1 = 0.7 * 2 * 10. Solving for x1, we get x1 = -0.875 meters. Since x represents displacement, it is negative to indicate compression of the spring.
Therefore, the minimum distance x1 we have to compress the spring for the block to start moving when released is 0.875 meters.
(b) To find the distance |x2 - x1| between the point of release x1 found in (a) and the point x2 where the block will come to a stop again, we need to take into account the forces acting on the block.
When the block is moving back towards x1, it experiences a kinetic friction force, opposed to the block's motion. The kinetic friction force is given by μk * N.
The force provided by the spring opposes the kinetic friction force, so we have -k * x2 = μk * m * g.
Plugging in the given values, we have -16 * x2 = 0.5 * 2 * 10. Solving for x2, we get x2 = -0.625 meters.
Therefore, the distance |x2 - x1| between the point of release x1 and the point x2 where the block will come to a stop again is 0.625 meters.
(c) To find the time t12 it takes the block to come to a rest after the release, we can use the equation of motion.
The equation of motion for an object undergoing simple harmonic motion is given by x = A * cos(ω * t), where x is the displacement, A is the amplitude, ω is the angular frequency, and t is the time.
In this case, since the block is coming to rest after the release, the displacement x is equal to x2 - x1.
The angular frequency ω is given by ω = √(k / m), where k is the spring constant and m is the mass of the block.
Using the equation x = A * cos(ω * t), we set x = x2 - x1 and A = (x2 - x1), since the amplitude is equal to the maximum displacement.
Rearranging the equation, we have (x2 - x1) * cos(ω * t) = x2 - x1.
Simplifying, cos(ω * t) = 1.
Taking the inverse cosine on both sides, we have ω * t = 0.
Since the cosine function repeats after a period of 2π, we can determine that ω * t should be an integer multiple of 2π. Therefore, ω * t = 2π.
Substituting ω = √(k / m), we have √(k / m) * t = 2π.
Simplifying, t = (2π * √(m / k)).
Plugging in the given values, we have t = (2π * √(2 / 16)) = 0.785 seconds.
Therefore, the time t12 it takes the block to come to a rest after the release is 0.785 seconds.
(d) After the block has come to a rest at point x2, it will move back towards x1 and oscillate with constant frequency and exponentially decreasing amplitude.
The correct answer is: The block will move back towards x1, and it will oscillate with constant frequency and exponentially decreasing amplitude.