Problem 30 - College Physics

Question

\(\bullet\) A 2.00 kg frictionless block is attached to an ideal spring with force constant 315 \(\mathrm{N} / \mathrm{m}\) . Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 \(\mathrm{m} / \mathrm{s} .\) Find (a) the amplitude of the motion, (b) the maximum acceleration of the block, and (c) the maximum force the spring exerts on the block.

Step-by-Step Solution

Answer

(a) A = 0.954 m, (b) a = 150.345 m/s², (c) F = 300.51 N.

Step 1: Calculate Amplitude Using Conservation of Energy

In a spring-mass system, mechanical energy is conserved. Initially, all energy is kinetic.The total mechanical energy (\(E\)) is:\[ E = \frac{1}{2} mv^2 \]where\( m = 2.00 \text{ kg} \)and\( v = 12.0 \text{ m/s} \)Substitute the values:\[ E = \frac{1}{2} (2.00) (12.0)^2 = \frac{1}{2} (2.00) (144) = 144 \text{ J} \]This energy is shared with the potential energy at maximum compression/stretching, where velocity is zero and amplitude is maximum:\[ E = \frac{1}{2} k A^2 \]Solve for amplitude\( A \):\[ 144 = \frac{1}{2} (315) A^2 \rightarrow 288 = 315 A^2 \]\[ A = \sqrt{\frac{288}{315}} \approx 0.954 \text{ m} \]

Step 2: Determine Maximum Acceleration

Maximum acceleration occurs at the maximum displacement (amplitude) when the spring force is greatest.The spring force is given by\( F = kx \), where\( x = A \). From Newton's second law,\( F = ma \), thus:\[ ma = kx \rightarrow a = \frac{kx}{m} \]Use\( x = A = 0.954 \text{ m} \), then\[ a = \frac{315 \times 0.954}{2} \approx 150.345 \text{ m/s}^2 \]

Step 3: Calculate Maximum Force Exerted by the Spring

Maximum force exerted by the spring occurs when the spring is at maximum displacement (or amplitude).The force is calculated using\( F = kx \), with\( x = A \):\[ F = 315 \times 0.954 \approx 300.51 \text{ N} \]

Key Concepts

Conservation of Energy

In a spring-mass system, the principle of conservation of energy plays a crucial role. The total mechanical energy (which is the sum of kinetic and potential energy) remains constant in the absence of non-conservative forces like friction. Initially, when the block is just released, it possesses all its energy as kinetic energy because its speed is maximum and the spring is neither compressed nor stretched. The initial kinetic energy of the block is given by:\[E_k = \frac{1}{2}mv^2\]where \(m\) is the mass of the block and \(v\) is its velocity. This energy is later converted to potential energy as the spring stretches or compresses. At maximum displacement, all the kinetic energy transforms into potential energy:\[E_p = \frac{1}{2}kA^2\]Here, \(k\) is the spring constant, and \(A\) is the amplitude, or the maximum displacement. By setting the equations for kinetic and potential energy equal at the two scenarios, we can solve for the amplitude.

Spring-Mass System

A spring-mass system consists of masses connected by springs. For this exercise, we assume an ideal spring with no mass and no energy loss due to damping or resistance. The spring's force constant, or spring constant \(k\), determines the stiffness of the spring. The higher the spring constant, the stiffer the spring.

• Force exerted by the spring is proportional to the displacement \(x\) from its equilibrium position, given by Hooke’s law: \(F = -kx\).
• The negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement.

For our problem, when the spring is stretched or compressed to its maximum displacement, the energy conversion takes place smoothly, demonstrating the ideal behavior of springs.

Maximum Displacement

Maximum displacement or amplitude in a spring-mass system is the farthest point the mass reaches from its equilibrium position. At this point, the block momentarily comes to rest, and all the kinetic energy has been transformed into potential energy stored in the spring. To find the maximum displacement, we use the fact that all the kinetic energy is converted to potential energy:

• Energy conservation equation: \( \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \).
• The amplitude \(A\) can be solved by rearranging the conservation of energy formula.

In this scenario, the spring force reaches its peak, exerting the maximum force on the block, but the speed is zero.

Newton's Second Law

Newton’s Second Law is fundamental in understanding the dynamics of the spring-mass system. It states that the force acting on an object is equal to the mass of the object times its acceleration \(F = ma\). In the context of our problem:

• The maximum acceleration occurs when the spring is at maximum displacement. At this point, the acceleration due to the spring force is maximum.
• Since the force from the spring at maximum displacement is given by \(F = kx\) (where \(x\) is the maximum displacement), we can use \(F = ma\) to find acceleration: \(a = \frac{kx}{m}\).

By plugging in the known values for the spring constant and the mass, we derive the maximum acceleration the block experiences within this system.