Let's solve each part step-by-step: --- ## (a) General Energy Equation The **general equation** linking any two energy states of a mechanical system (neglecting energy losses) is: \[ T_1 + V_1 + W_{1 \to 2} = T_2 + V_2 \] Where: - \(T\) = Kinetic energy - \(V\) = Potential energy (gravitational + elastic) - \(W_{1 \to 2}\) = Work done by non-conservative forces (zero here) For a conservative system: \[ T_1 + V_1 = T_2 + V_2 \] --- ## (b) Initial Energy of the System **Datum:** Take the spring’s uncompressed top as the reference level for gravitational potential energy. ### **Given:** - Mass, \(m = 5\,\mathrm{kg}\) - Spring constant, \(k = 600\,\mathrm{N/m}\) - Pre-compression, \(\delta_ = 40\,\mathrm{mm} = .04\,\mathrm{m}\) - Height above spring, \(h = 80\,\mathrm{mm} = .08\,\mathrm{m}\) ### **Initial State (Before Release):** - Cylinder: Stationary, held 80 mm above spring - Spring: Pre-compressed by 40 mm ### **Energies:** - **Kinetic Energy (\(T_1\))**: \(\) (cylinder at rest) - **Gravitational Potential Energy (\(V_{g1}\))**: \(mgh\) - **Spring Potential Energy (\(V_{s1}\))**: \(\frac{1}{2}k\delta_^2\) #### **Total Initial Energy (\(E_{initial}\)):** \[ E_{initial} = T_1 + V_{g1} + V_{s1} = + mg h + \frac{1}{2} k \delta_^2 \] \[ E_{initial} = (5)(9.81)(.08) + \frac{1}{2}(600)(.04)^2 \] \[ = (5 \times 9.81 \times .08) + (.5 \times 600 \times .0016) \] \[ = 3.924 + 4.8 = \boxed{8.72\,\mathrm{J}} \] --- ## (c) Maximum Compression of the Spring (\(x\)) Let \(x\) = **additional compression** after impact (beyond pre-compression). **At maximum compression:** - Cylinder’s velocity = (all energy is potential) - All energy is stored in compressed spring and gravitational potential (now at new lower position). Let total compression from spring's free length = \(x_{max}\). - Initial pre-compression: \(\delta_\) - Additional compression: \(x\) - So, \(x_{max} = \delta_ + x\) ### **Energy Conservation:** \[ E_{initial} = \frac{1}{2}k x_{max}^2 + mg(-x) \] But we need to set all reference heights properly: - Cylinder drops \(h\) to contact, then compresses spring by \(x\). - The change in gravitational PE during compression = \(-mg x\) - So, at max compression: \[ E_{initial} + mg \cdot = \frac{1}{2}k(\delta_ + x)^2 + mg(-x) \] But initial gravitational PE is at \(h\) above pre-compressed position. So, \[ mg h + \frac{1}{2}k\delta_^2 = \frac{1}{2}k(\delta_ + x)^2 - mgx \] **Expand:** \[ mg h + \frac{1}{2}k\delta_^2 = \frac{1}{2}k(\delta_^2 + 2\delta_ x + x^2) - mgx \] \[ mg h + \frac{1}{2}k\delta_^2 = \frac{1}{2}k\delta_^2 + k\delta_ x + \frac{1}{2}k x^2 - mgx \] \[ mg h = k\delta_ x + \frac{1}{2}k x^2 - mgx \] \[ mg h = k\delta_ x + \frac{1}{2}k x^2 - mgx \] \[ mg h = k\delta_ x + \frac{1}{2}k x^2 - mgx \] Bring terms to one side: \[ mg h = k\delta_ x + \frac{1}{2}k x^2 - mgx \] \[ mg h = k\delta_ x + \frac{1}{2}k x^2 - mgx \] \[ mg h + mgx = k\delta_ x + \frac{1}{2}k x^2 \] \[ mg(h + x) = k\delta_ x + \frac{1}{2}k x^2 \] Now, substitute values: - \(mg = 5 \times 9.81 = 49.05\,\mathrm{N}\) - \(k = 600\,\mathrm{N/m}\) - \(\delta_ = .04\,\mathrm{m}\) - \(h = .08\,\mathrm{m}\) \[ 49.05(.08 + x) = 600 \times .04 x + 300 x^2 \] \[ 49.05 \times .08 + 49.05 x = 240 x + 300 x^2 \] \[ 3.924 + 49.05 x = 240 x + 300 x^2 \] \[ 3.924 = 240 x + 300 x^2 - 49.05 x \] \[ 3.924 = (240 - 49.05)x + 300 x^2 \] \[ 3.924 = 190.95 x + 300 x^2 \] **Quadratic Equation:** \[ 300 x^2 + 190.95 x - 3.924 = \] **Use quadratic formula:** \[ x = \frac{ -190.95 \pm \sqrt{(190.95)^2 - 4 \times 300 \times (-3.924)} }{2 \times 300} \] Calculate discriminant: \[ (190.95)^2 = 36462 \] \[ 4 \times 300 \times 3.924 = 47088 \] \[ \text{Discriminant} = 36462 + 47088 = 83550 \] \[ \sqrt{83550} \approx 289 \] Now, \[ x = \frac{ -190.95 \pm 289 }{600} \] Take the positive root: \[ x = \frac{ -190.95 + 289 }{600} = \frac{98.05}{600} = .0163\,\mathrm{m} = 16.3\,\mathrm{mm} \] **Maximum compression of spring:** \[ \boxed{16.3\,\mathrm{mm}} \] --- ## (d) Impulse Exerted by the Spring Impulse, \(J\), is the change in momentum: \[ J = \Delta p = m(v_{before} - v_{after}) \] - \(v_{after} = \) (comes to rest) - \(v_{before}\): Just before impact First, **find velocity just before impact** (drop from \(h = .08\,\mathrm{m}\)): By energy: \[ mgh = \frac{1}{2}mv^2 \] \[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times .08} \approx \sqrt{1.5696} \approx 1.25\,\mathrm{m/s} \] So, \[ J = m(v_{before} - v_{after}) = 5 \times (1.25 - ) = 6.25\,\mathrm{N\,s} \] However, **gravitational force also exerts an impulse** during contact time (\(\Delta t = .026\,\mathrm{s}\)): Impulse due to gravity during contact: \[ J_g = mg \Delta t = 49.05 \times .026 = 1.275\,\mathrm{N\,s} \] **Total impulse by spring:** \[ J_{spring} = J + J_g = 6.25 + 1.275 = \boxed{7.53\,\mathrm{N\,s}} \] --- ## (e) Average Contact Force During Impact \[ \text{Average Force}, \quad F_{avg} = \frac{\text{Impulse}}{\Delta t} \] \[ F_{avg} = \frac{7.53}{.026} \approx 289.8\,\mathrm{N} \] \[ \boxed{290\,\mathrm{N}} \] --- # **Summary of Answers** **(a)** \(T_1 + V_1 = T_2 + V_2\) (Conservation of Mechanical Energy) **(b)** Initial energy of the system: \(\boxed{8.72\,\mathrm{J}}\) **(c)** Maximum additional compression of the spring: \(\boxed{16.3\,\mathrm{mm}}\) **(d)** Magnitude of impulse exerted by the spring: \(\boxed{7.53\,\mathrm{N\,s}}\) **(e)** Average contact force during impact: \(\boxed{290\,\mathrm{N}}\)