Two identical 0.18 kg blocks (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper diagram, where x2 = 0.06 m. Then a constant force of 9 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower diagram, where x1 = 0.01 m and x3 = 0.1 m. At this final time, the system is moving to the right and also vibrating, and the spring is stretched.
What is the spring constant of the spring?
The spring constant of the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is equal to the spring constant multiplied by the displacement of the spring. In this case, the displacement of the spring is x3 - x2 = 0.1 m - 0.06 m = 0.04 m. Therefore, the spring constant can be calculated as:
Spring Constant = Force / Displacement = 9 N / 0.04 m = 225 N/m
To find the equation of motion for each block, we can first use Newton's second law for each block separately.
For Block 1:
The net force acting on Block 1 is due to the spring and the applied force. We can write the equation of motion as:
m1 * a1 = -k * x1 + k * (x2 - x1) + F_applied
where m1 is the mass of Block 1, a1 is the acceleration of Block 1, k is the spring constant, x1 is the displacement of Block 1 from its equilibrium position, x2 is the displacement of Block 2 from its equilibrium position, and F_applied is the applied force.
For Block 2:
The net force acting on Block 2 is due to the spring and the applied force. We can write the equation of motion as:
m2 * a2 = -k * (x2 - x1) - F_applied
where m2 is the mass of Block 2, a2 is the acceleration of Block 2, k is the spring constant, x1 is the displacement of Block 1 from its equilibrium position, x2 is the displacement of Block 2 from its equilibrium position, and F_applied is the applied force.
Since the system is at rest initially and moving to the right at a later time, we know that Block 2 has a nonzero acceleration while Block 1 has a negligible acceleration compared to the applied force. Thus, we can neglect Block 1's acceleration and simplify the equation of motion for Block 1 to:
0 = k * (x2 - x1) + F_applied
To find the spring constant, we can use Hooke's law. When the spring is stretched, the force exerted by the spring is proportional to the displacement. We can write this as:
F_spring = k * (x2 - x1)
Given that the spring force is 9 N when the displacement is 0.06 m - 0.01 m = 0.05 m, we can substitute these values and solve for the spring constant:
9 N = k * 0.05 m
Now we can substitute the value of the spring constant into the equation of motion for Block 1 to find the applied force:
0 = k * (0.06 m - 0.01 m) + F_applied
0 = 9 N - F_applied
F_applied = 9 N
Therefore, the applied force is 9 N.
To understand the situation and answer the question, we need to apply the principles of physics, specifically Newton's laws of motion and Hooke's law for springs.
First, let's analyze the initial position. We have two identical blocks, labeled 1 and 2, connected by a spring. Block 1 is on the left with position x1, and block 2 is on the right with position x2. The spring is initially unstretched, so the spring force is zero (Fs = 0).
Now, a constant force of 9 N is applied to block 2. As a result, the system will experience an external force (Fe) acting on block 2 in the rightward direction.
At the final time, the blocks have moved to new positions. Block 1 is now at position x1 = 0.01 m, and block 2 is at position x2 = 0.06 m. The spring is stretched, so there is a spring force (Fs) acting on block 2 in the leftward direction.
We are asked about the force exerted by the spring on block 2 at this final time. To find this force, we can apply Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement.
F = -kΔx
Where F is the force exerted by the spring, k is the spring constant, and Δx is the displacement from the equilibrium position.
In our case, Δx is the difference between the final position of block 2 (x2 = 0.06 m) and the equilibrium position (x2 = 0 m). So, Δx = 0.06 m - 0 m = 0.06 m.
Now, we need to determine the spring constant, k. The spring constant depends on the properties of the spring and can be calculated using Hooke's law with known values of force and displacement.
Given that the spring force is zero when the spring is unstretched, we know that the force exerted by the spring when it is stretched by Δx = 0.06 m is equal to the external force applied to block 2, which is Fe = 9 N.
So, using Hooke's law:
9 N = -k * 0.06 m
To find k, we isolate it:
k = -9 N / 0.06 m
Calculating this expression, we find:
k ≈ 150 N/m
Now that we have the spring constant (k) and the displacement (Δx = 0.06 m), we can calculate the force exerted by the spring on block 2:
F = -kΔx
F = -(150 N/m)(0.06 m)
F ≈ -9 N
Therefore, the force exerted by the spring on block 2 at the final time is approximately -9 N. Note that the negative sign indicates that the force is in the opposite direction to the displacement.