# Problem Overview A 40 kg block moves along an inclined ramp with an attached spring The problem involves the maximum extension the spring and work done by various forces the block's motion. --- # Data | | Value | | |------------|--------|--------------| Mass of | \( m 40\, \text{kg \) || Incline | \( \theta = 30circ \) | | Unretched spring position | \( s_ = .10\, \{m}) | measured the ramp | | Coefficient of and kinetic friction | \( \mu_s = \mu_k = .20 \) | | | Initial position | \( s_i = .60\, \text{m} \) | | | Spring constant | \( k = 100\, \text{N/m} \) | | --- # Part (a): Maximum Distance \( s_{\text{max}} \) **Goal:** Find the maximum extension \( s_{\text{max}} \) of the spring along the ramp, measured from the origin. --- ## Step 1: Convert positions to energies - Initial position: \( s_i = .60\, \text{m} \) - Spring is at rest at \( s_ = .10\, \text{m} \) --- ## Step 2: Write energy considerations At maximum extension, the block comes to rest, so kinetic energy is zero. Total mechanical energy at initial position (considering work done by non-conservative forces): \[ E_{\text{initial}} = U_g(i) + U_s(i) + W_{\text{friction}}(i \to s_{\text{max}}) \] At maximum extension: \[ E_{\text{final}} = U_g(f) + U_s(f) + \] since the block stops, kinetic energy is zero. --- ## Step 3: Calculate gravitational potential energy change \[ U_g = m g h \] with: \[ h = s \sin \theta \] Change in gravitational potential energy: \[ \Delta U_g = m g (s \sin \theta - s_i \sin \theta) = m g \sin \theta (s - s_i) \] --- ## Step 4: Spring potential energy \[ U_s = \frac{1}{2} k (s - s_)^2 \] Change in spring potential energy: \[ \Delta U_s = \frac{1}{2} k (s - s_)^2 - \frac{1}{2} k (s_i - s_)^2 \] --- ## Step 5: Friction work Friction opposes motion, doing negative work: \[ W_{\text{friction}} = -f_{\text{fric}} \times \text{distance} \] where: \[ f_{\text{fric}} = \mu_k N \] Normal force: \[ N = m g \cos \theta \] since the component of weight normal to ramp. Total work done by friction: \[ W_f = - \mu_k N (s - s_i) = - \mu_k m g \cos \theta (s - s_i) \] --- ## Step 6: Write energy balance \[ = \big[ m g \sin \theta (s - s_i) \big] + \big[ \frac{1}{2} k (s - s_)^2 - \frac{1}{2} k (s_i - s_)^2 \big] + \big[ - \mu_k m g \cos \theta (s - s_i) \big] \] which simplifies to: \[ = (m g \sin \theta - \mu_k m g \cos \theta)(s - s_i) + \frac{1}{2} k (s - s_)^2 - \frac{1}{2} k (s_i - s_)^2 \] --- ## Step 7: Plug in known values Calculate constants: \[ g = 9.8\, \text{m/s}^2 \] \[ A = m g (\sin \theta - \mu_k \cos \theta) \] \[ A = 40 \times 9.8 \times (\sin 30^\circ - .2 \times \cos 30^\circ) \] \[ \sin 30^\circ = .5,\quad \cos 30^\circ \approx .866 \] \[ A = 392 \times (.5 - .2 \times .866) = 392 \times (.5 - .1732) = 392 \times .3268 \approx 128. \] Similarly, compute the spring energy difference: \[ B = \frac{1}{2} \times 100 \times [(s - .10)^2 - (.60 - .10)^2] \] \[ = 50 \times [(s - .10)^2 - .50^2] = 50 \times [(s - .10)^2 - .25] \] --- ## Step 8: Final quadratic in \( s \) Set the energy equation: \[ A (s - .60) + B = \] which becomes: \[ 128. (s - .60) + 50 [(s - .10)^2 - .25] = \] --- ## Step 9: Solve for \( s \) Expand: \[ 128. (s - .60) + 50 (s - .10)^2 - 12.5 = \] \[ 128 s - 76.8 + 50 (s^2 - .2 s + .01) - 12.5 = \] \[ 128 s - 76.8 + 50 s^2 - 10 s + .5 - 12.5 = \] Combine like terms: \[ 50 s^2 + (128 s - 10 s) + (-76.8 + .5 - 12.5) = \] \[ 50 s^2 + 118 s - 88.8 = \] --- ## Step 10: Quadratic formula \[ s = \frac{-118 \pm \sqrt{(118)^2 - 4 \times 50 \times (-88.8)}}{2 \times 50} \] Calculate discriminant: \[ D = 118^2 - 4 \times 50 \times (-88.8) = 13924 + 17760 = 31684 \] \[ \sqrt{D} \approx 178. \] Find roots: \[ s = \frac{-118 \pm 178}{100} \] Two solutions: - \( s = \frac{-118 + 178}{100} = \frac{60}{100} = .60\, \text{m} \) - \( s = \frac{-118 - 178}{100} = \frac{-296}{100} = -2.96\, \text{m} \) (discard, physically impossible) --- ## **Answer for (a):** \[ \boxed{ \textbf{Maximum extension } s_{\text{max}} \approx .60\, \text{m} } \] which is the same as the initial position \( s_i \). This indicates the block reaches the maximum extension at the initial position, implying it just stops at \( s = .60\, \text{m} \). --- # Part (b): Work Done by Forces ## (i) Work by gravity: \[ W_g = m g \sin \theta (s - s_i) \] from \( s_i = .60\, \text{m} \) to \( s_{\text{max}} \): \[ W_g = 128. \times (.60 - .60) = \] **No net work** since the start and end positions are the same, but if the block moved further, this would be relevant. ## (ii) Work by spring: \[ W_s = \frac{1}{2} k (s_{\text{max}} - s_)^2 - \frac{1}{2} k (s_i - s_)^2 \] \[ = 50 \times (.60 - .10)^2 - 50 \times (.60 - .10)^2 = \] since initial and maximum positions are the same, the work done by the spring is zero at the stop point. ## (iii) Work by friction: \[ W_f = - \mu_k m g \cos \theta (s_{\text{max}} - s_i) = -128. \times (.60 - .60) = \] --- # **Summary** - The block reaches a maximum extension of approximately **.60 m** along the ramp. - The total work done by all forces during the motion from \( s_ \) to \( s_i \) sums to zero since the block ends at rest at the same position it started (no further movement). --- # **Note:** The calculations suggest that the block just reaches the initial position, indicating the initial release at \( s_i=.60\, \text{m} \) is at the turning point. If more precise or different initial conditions are considered, the method applies similarly.