Question

A block (mass \(m\) ) hangs from a spring (spring constant k). The block is released from rest a distance \(d\) above its equilibrium position. (a) What is the speed of the block as it passes through the equilibrium point? (b) What is the maximum distance below the equilibrium point that the block will reach?

Short answer

Question: A block is attached to a spring and held at a distance \(d\) above its equilibrium position. The block is then released from rest. Find the (a) speed of the block as it passes through the equilibrium position and (b) the maximum distance the block will travel below its equilibrium position. Answer: (a) The speed of the block as it passes through the equilibrium position is \(v = \sqrt{\frac{kd^2}{m}}\). (b) The maximum distance below the equilibrium position the block will travel is \(d_{max} = \sqrt{2d^2}\).

Step-by-step solution

(Part a: Identifying the Initial and Final Conditions)

Before solving for the speed, let's identify the initial and final conditions: 1. Initial position: block is held at a distance d above its equilibrium position. 2. Final position: block passes through equilibrium position. Now we can apply the principle of conservation of mechanical energy: the initial mechanical energy (spring potential energy) equals the final mechanical energy (sum of kinetic and potential energies) at the final position.

(Part a: Spring Potential Energy and Conservation of Energy)

The spring potential energy is given by the formula: \(U_s=\frac{1}{2}kd^2\) Where \(k\) is the spring constant and \(d\) is the distance displaced from equilibrium. Since mechanical energy is conserved, the initial potential energy of the spring equals the sum of kinetic energy and potential energy of the block at the equilibrium position: \(U_s = K_f + U_f\) As the block passes through the equilibrium position, the potential energy will be zero:

(Part a: Solving for the Final Kinetic Energy and Speed)

Therefore, we have: \(U_s = K_f\) \(\frac{1}{2}kd^2 = \frac{1}{2}mv^2\), where \(v\) is the speed of the block as it passes through equilibrium. Solving for \(v\): \(v = \sqrt{\frac{kd^2}{m}}\)

(Part b: Maximum Compression and Equating Potential and Kinetic Energies)

Now we need to find the maximum distance below the equilibrium point that the block will reach. As the block moves downwards, its speed will decrease due to the increasing spring force, and eventually, it will come to a halt for an instant before moving upward. At this point, all its kinetic energy will have been converted into spring potential energy. Let's call the maximum distance below equilibrium position \(d_{max}\). We can write the conservation of energy equation at this instant: \(U_{s,initial} + K_{initial} = U_{s,max\_compression} + K_{max\_compression}\) Note that \(K_{initial}\) is zero since the block is released from rest, and that \(K_{max\_compression}\) is also zero as the block momentarily comes to a halt.

(Part b: Solving for Maximum Distance Below Equilibrium Position)

The above equation simplifies to: \(U_{s,initial} = U_{s,max\_compression}\) \(\frac{1}{2}kd^2 = \frac{1}{2}k(d_{max})^2\) Solving for \(d_{max}\): \(d_{max} = \sqrt{2d^2}\) So, the solutions are: (a) Speed of the block as it passes through the equilibrium point: \(v = \sqrt{\frac{kd^2}{m}}\) (b) Maximum distance below the equilibrium point: \(d_{max} = \sqrt{2d^2}\)