# Conservation of Energy Problem: Updated Solution (with \( k = 520\,\mathrm{N/m} \)) Let's redo the calculation by replacing the spring constant \( k = 370\,\mathrm{N/m} \) with \( k = 520\,\mathrm{N/m} \). ## 1. Given Data - Mass, \( m = .5\,\mathrm{kg} \) - Gravity, \( g = 9.81\,\mathrm{m/s^2} \) - Unstretched length of spring, \( L_ = .2\,\mathrm{m} \) - \( AB: \) \( x = .3\,\mathrm{m}, y = .15\,\mathrm{m} \) - \( AC: \) \( x = .3\,\mathrm{m}, y = .075\,\mathrm{m} \) - Spring constant, \( k = 520\,\mathrm{N/m} \) --- ## 2. Length Calculations ### \( L_{AB} \): \[ L_{AB} = \sqrt{ (.3)^2 + (.15)^2 + (.075)^2 } = \sqrt{ .09 + .0225 + .005625 } = \sqrt{ .118125 } = .3437\,\mathrm{m} \] ### \( L_{AC} \): \[ L_{AC} = \sqrt{ (.3)^2 + (.075)^2 } = \sqrt{ .09 + .005625 } = \sqrt{ .095625 } = .3093\,\mathrm{m} \] --- ## 3. Spring Elongation ### At B \[ \Delta L_{AB} = L_{AB} - L_ = .3437 - .2 = .1437\,\mathrm{m} \] ### At C \[ \Delta L_{AC} = L_{AC} - L_ = .3093 - .2 = .1093\,\mathrm{m} \] --- ## 4. Energies ### At B (Rest) - **Spring Energy:** \[ E_{1} = \frac{1}{2} k (\Delta L_{AB})^2 = \frac{1}{2} \times 520 \times (.1437)^2 = 260 \times .02066 = 5.3716\,\mathrm{J} \] - **Potential Energy from weight (height = .15):** \[ E_{2} = mgh = .5 \times 9.81 \times .15 = .73575\,\mathrm{J} \] - **Total at B:** \[ E_B = E_1 + E_2 = 5.3716 + .73575 = 6.1073\,\mathrm{J} \] ### At C (Velocity = \( v_C \)) - **Kinetic Energy:** \[ T_C = \frac{1}{2} m v_C^2 = .25 v_C^2 \] - **Spring Energy:** \[ E_C = \frac{1}{2} k (\Delta L_{AC})^2 = \frac{1}{2} \times 520 \times (.1093)^2 = 260 \times .011943 = 3.1052\,\mathrm{J} \] --- ## 5. Conservation of Energy \[ E_B = T_C + E_C \] \[ 6.1073 = .25 v_C^2 + 3.1052 \] \[ .25 v_C^2 = 6.1073 - 3.1052 = 3.0021 \] \[ v_C^2 = \frac{3.0021}{.25} = 12.0084 \] \[ v_C = \sqrt{12.0084} = 3.465\,\mathrm{m/s} \] --- ## 6. Force of Rod on Collar at C ### Direction Angles \[ \theta = \tan^{-1}\left(\frac{.075}{.3}\right) = \tan^{-1}(.25) = 14.04^\circ \] ### Spring Force Magnitude \[ F_S = k \Delta L_{AC} = 520 \times .1093 = 56.836\,\mathrm{N} \] ### Components \[ F_{Si} = F_S \cos\theta = 56.836 \times \cos(14.04^\circ) = 55.151\,\mathrm{N} \] \[ F_{Sk} = F_S \sin\theta = 56.836 \times \sin(14.04^\circ) = 13.792\,\mathrm{N} \] ### Net Forces Sum of forces at C: \[ \sum F = (F_x + F_{Si})\hat{i} + (F_y - mg)\hat{j} + F_{Sk}\hat{k} \] From dynamics: - Radial (j): \( \sum F_j = m \frac{v_C^2}{r} \) - Out of plane (k): \( \sum F_k = m a_k \) (if any acceleration component) #### Equate the \( \hat{i} \), \( \hat{j} \) components: - \( F_x + 55.151 = \implies F_x = -55.151\,\mathrm{N} \) - \( F_y - 4.905 = m \frac{v_C^2}{.15} \) \[ F_y = 4.905 + .5 \times \frac{(3.465)^2}{.15} = 4.905 + .5 \times 80.048 = 4.905 + 40.024 = 44.929\,\mathrm{N} \] --- ## **Final Answers** - **Speed at C:** \( \boxed{v_C = 3.47\,\mathrm{m/s}} \) - **Force by the rod on the collar at C:** \[ \boxed{ \vec{F} = -55.15\,\hat{i} + 44.93\,\hat{j}\ \mathrm{N} } \] --- ## **Summary Table** | Quantity | Value (rounded) | |----------------------|---------------------| | \( v_C \) | \( 3.47\,\mathrm{m/s} \) | | \( F_x \) | \( -55.15\,\mathrm{N} \) | | \( F_y \) | \( 44.93\,\mathrm{N} \) | --- ### **Notes** - All calculations use \( k = 520\,\mathrm{N/m} \). - All lengths and energies recalculated accordingly. - The negative sign in \( F_x \) indicates direction opposite to the \( \hat{i} \) axis. --- **If you need the vector in component form or want to see the full free-body diagram, let me know!**