Problem 5 - AP Physics 1 with 2 Practice Test 2nd

Question

A mass of \(0.3 \mathrm{~kg}\) is connected to a massless spring with a force constant \(k\) of \(20 \mathrm{~N} / \mathrm{m}\). The system oscillates horizontally on a frictionless surface with an amplitude of \(4 \mathrm{~cm}\). What is the velocity of the mass when it is \(2 \mathrm{~cm}\) from its equilibrium position? (A) \(0.28 \mathrm{~m} / \mathrm{s}\) (B) \(0.08 \mathrm{~m} / \mathrm{s}\) (C) \(0.52 \mathrm{~m} / \mathrm{s}\) (D) \(0.15 \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The velocity is \(0.28 \mathrm{~m/s}\). The correct answer is (A).

Step 1: Write down the given values

Mass of the object, \(m = 0.3 \mathrm{~kg}\). Force constant, \(k = 20 \mathrm{~N} / \mathrm{m}\). Amplitude of oscillation, \(A = 4 \mathrm{~cm} = 0.04 \mathrm{~m}\). Position from equilibrium, \(x = 2 \mathrm{~cm} = 0.02 \mathrm{~m}\).

Step 2: Expression for total mechanical energy in the system

The total mechanical energy (E) in a spring-mass system is given by: \[ E = \frac{1}{2} k A^2 \]

Step 3: Expression for potential energy at any displacement

The potential energy (U) at position \(x\) is: \[ U = \frac{1}{2} k x^2 \]

Step 4: Calculate the total mechanical energy

Substitute \(k = 20 \mathrm{~N} / \mathrm{m}\) and \(A = 0.04 \mathrm{~m}\) into the formula: \[ E = \frac{1}{2} (20) (0.04)^2 \] \[ E = 0.016 \mathrm{~J} \]

Step 5: Calculate the potential energy at position x

Substitute \(k = 20 \mathrm{~N} / \mathrm{m}\) and \(x = 0.02 \mathrm{~m}\) into the formula: \[ U = \frac{1}{2} (20) (0.02)^2 \] \[ U = 0.004 \mathrm{~J} \]

Step 6: Expression for kinetic energy at position x

The kinetic energy (K) at position \(x\) is the remaining energy: \[ K = E - U \] Substitute \(E = 0.016 \mathrm{~J}\) and \(U = 0.004 \mathrm{~J}\): \[ K = 0.016 - 0.004 \] \[ K = 0.012 \mathrm{~J} \]

Step 7: Expression for velocity from kinetic energy

The kinetic energy is also given by: \[ K = \frac{1}{2} m v^2 \] Solve for \(v\): \[ v = \frac{v}{m} \] \( v = \sqrt{\frac{2K}{m}} \)

Step 8: Calculate the velocity

Substitute \(K = 0.012 \mathrm{~J}\) and \(m = 0.3 \mathrm{~kg}\) into the formula: \[ v = \sqrt{\frac{2 (0.012)}{0.3}} \] \[ v = \sqrt{0.08} \] \[ v = 0.28 \mathrm{~m/s} \]

Key Concepts

Oscillating Systems

An oscillating system is one where an object moves back and forth around an equilibrium position. A classic example is a mass attached to a spring. This type of motion is called simple harmonic motion (SHM). SHM occurs when the force acting on the object is directly proportional to its displacement from the equilibrium position and acts in the opposite direction. This restoring force drives the oscillation.

In the problem, we have a mass of 0.3 kg attached to a massless spring on a frictionless surface. This setup is an ideal example of a horizontal oscillating system. The mass oscillates around an equilibrium position with a certain amplitude, in this case, 4 cm. No friction means no energy loss, which is crucial for truly simple harmonic motion to occur.

Some key points about oscillating systems include:

• They have a natural frequency determined by the mass and spring constant.
• The total mechanical energy is conserved if there's no friction.
• The motion can be described by sinusoidal functions for displacement, velocity, and acceleration.

Spring Constant

The spring constant, denoted as 𝑘, measures the stiffness of the spring. It is given in units of Newtons per meter (N/m). A higher spring constant means a stiffer spring and a stronger restoring force for a given displacement.

In our problem, the spring constant is 20 N/m. This is part of Hooke's Law, which states that the force exerted by the spring is proportional to the displacement: \( F = -kx \). This negative sign shows that the force acts in the opposite direction to the displacement, which helps in creating the oscillation.

The spring constant determines not just the force but also contributes to the total mechanical energy of the system. It combines with the amplitude to give us the total amount of energy stored in the system when the spring is at maximum displacement:

• Total Mechanical Energy: \[ E = \frac{1}{2} k A^2 \]

This energy is transferred back and forth between kinetic and potential forms as the mass oscillates.

Kinetic Energy

Kinetic energy (KE) in an oscillating system is the energy due to the mass's motion. When the spring is at its equilibrium position, all the system's energy is kinetic because the potential energy (due to the spring's compression or elongation) is zero.

As the mass moves toward the equilibrium from either side, it speeds up, converting potential energy (PE) into kinetic energy. The kinetic energy at any position can be found using:

• \[ K = \frac{1}{2} mv^2 \]
• The velocity \( v \) can be derived from knowing the total mechanical energy and the instantaneous potential energy: \[ v = \sqrt{\frac{2K}{m}} \]

In our example, when the mass is 2 cm from equilibrium, part of the total mechanical energy is still potential energy, and the remainder is kinetic energy. We calculated the kinetic energy at that position and then found the velocity by rearranging the kinetic energy formula. This lets us determine how fast the mass is moving at any given point in its cycle.

Understanding kinetic energy in simple harmonic motion is crucial. It tells us how the energy transitions and helps predict the object's velocity. This knowledge is essential for various applications, from clocks to engineering systems.