Questions: A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N / m. Initially the spring (whose other end is fixed) is stretched by 0.100 m and the attached glider is moving at 0.400 m / s in the direction that causes the spring to stretch farther. What is the total mechanical energy (kinetic energy plus elastic potential energy) of the system? Express your answer with the appropriate units. When the glider comes momentarily to rest by distance is the spring stretched? Express your answer with the appropriate units. x= Value Units Units input for part B

A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N / m. Initially the spring (whose other end is fixed) is stretched by 0.100 m and the attached glider is moving at 0.400 m / s in the direction that causes the spring to stretch farther. What is the total mechanical energy (kinetic energy plus elastic potential energy) of the system?

Express your answer with the appropriate units.

When the glider comes momentarily to rest by distance is the spring stretched?
Express your answer with the appropriate units.

x=
Value
Units
Units input for part B

Transcript text: A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant $6.00 \mathrm{~N} / \mathrm{m}$. Initially the spring (whose other end is fixed) is stretched by 0.100 m and the attached glider is moving at $0.400 \mathrm{~m} / \mathrm{s}$ in the direction that causes the spring to stretch farther. What is the total mechanical energy (kinetic energy plus elastic potential energy) of the system? Express your answer with the appropriate units. $\square$ Submit Request Answer Part B When the glider comes momentarily to rest by distance is the spring stretched? Express your answer with the appropriate units. $\square$ \[ \begin{array}{l} x= \\ \text { Value } \\ \text { Units } \\ \text { Units input for part B } \end{array} \] Submit Request Answer

Solution

Solution Steps

Step 1: Calculate the Initial Kinetic Energy

The kinetic energy (KE) of the glider can be calculated using the formula: \[ \text{KE} = \frac{1}{2} m v^2 \] where $ m = 0.240 \, \text{kg} $ is the mass of the glider and $ v = 0.400 \, \text{m/s} $ is its velocity.

\[ \text{KE} = \frac{1}{2} \times 0.240 \, \text{kg} \times (0.400 \, \text{m/s})^2 = 0.0192 \, \text{J} \]

Step 2: Calculate the Initial Elastic Potential Energy

The elastic potential energy (PE) stored in the spring is given by: \[ \text{PE} = \frac{1}{2} k x^2 \] where $ k = 6.00 \, \text{N/m} $ is the spring constant and $ x = 0.100 \, \text{m} $ is the initial stretch of the spring.

\[ \text{PE} = \frac{1}{2} \times 6.00 \, \text{N/m} \times (0.100 \, \text{m})^2 = 0.0300 \, \text{J} \]

Step 3: Calculate the Total Mechanical Energy

The total mechanical energy (E) of the system is the sum of the kinetic energy and the elastic potential energy: \[ E = \text{KE} + \text{PE} = 0.0192 \, \text{J} + 0.0300 \, \text{J} = 0.0492 \, \text{J} \]

Step 4: Determine the Maximum Stretch of the Spring

When the glider comes momentarily to rest, all the mechanical energy is converted into elastic potential energy. Thus, we set the total mechanical energy equal to the potential energy at maximum stretch: \[ E = \frac{1}{2} k x_{\text{max}}^2 \] Solving for $ x_{\text{max}} $: \[ 0.0492 \, \text{J} = \frac{1}{2} \times 6.00 \, \text{N/m} \times x_{\text{max}}^2 \] \[ x_{\text{max}}^2 = \frac{0.0492 \, \text{J}}{3.00 \, \text{N/m}} = 0.0164 \, \text{m}^2 \] \[ x_{\text{max}} = \sqrt{0.0164 \, \text{m}^2} = 0.1281 \, \text{m} \]