Question

A 0.428 kg mass is attached to a spring with a force constant of 26.4 N/m and released from rest a distance of 3.07 cm from the equilibrium position of the spring. Calculate the speed of the mass when it is halfway to the equilibrium position.

Answer 1

Answer 2

Calculation Steps:

1. Given Data:

• Mass ($m$) = 0.428 kg

• Spring constant ($k$) = 26.4 N/m

• Initial displacement ($x_{\text{max}}$) = 3.07 cm = 0.0307 m

• Halfway displacement ($x$) = $\frac{0.0307}{2}$ = 0.01535 m

• 
  

2. Total Energy of the System:
   The total energy is the potential energy stored in the spring at the initial displacement (since the mass is released from rest, kinetic energy is zero initially).

   $$E_{\text{total}}=\frac{1}{2}k{x}_{\text{max}}^2=\frac{1}{2}\times 26.4\times (0.0307{)}^2=0.01244\text{J}$$

4. Energy at Halfway Point:
   At halfway, the system has both potential energy and kinetic energy.

   $$E_{\text{total}}=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$$

   Plug in the values:

   $$0.01244=\frac{1}{2}\times 26.4\times (0.01535{)}^2+\frac{1}{2}\times 0.428\times v^2$$

   Simplify the potential energy term:

   $$\frac{1}{2}\times 26.4\times 0.0002356=0.00311\text{J}$$

   Now solve for kinetic energy:

   $$0.01244-0.00311=\frac{1}{2}\times 0.428\times v^2$$$$0.00933=0.214\times v^2$$

6. Solve for Speed ($v$):

   $$v^2=\frac{0.00933}{0.214}=0.0436$$$$v=\sqrt{0.0436}=0.418\text{m/s}$$

The mass is moving at 0.418 m/s when it is halfway to the equilibrium position