Let's solve the two required quantities step by step. --- ## Step 1: Convert Units - **Mass of the package**: \( 8) oz \( 1 \) lb = \( 16 \) oz \( m = \frac{8}{16} = .5 \) lb - **Weight**: \( W = mg \) \( g = 32.2 \) ft/s² \( W = .5 \) lb --- ## Step 2: Apply Conservation of Energy Let point A be at the bottom, and point C at the top (height \( h = 7.5 \) ft). - **At A**: - Kinetic Energy: \( \frac{1}{2} m v_^2 \) - Potential Energy: \( \) - **At C**: - Kinetic Energy: \( \frac{1}{2} m v_C^2 \) - Potential Energy: \( mgh \) Set the **total mechanical energy** at A equal to that at C: \[ \frac{1}{2} m v_^2 = \frac{1}{2} m v_C^2 + mgh \] \[ v_^2 = v_C^2 + 2gh \] --- ## Step 3: Minimum Speed at Top (Point C) At the top (point C), the **minimum speed** is when the normal force is zero (just about to lose contact): \[ \text{At C:} \quad \Sigma F = m \frac{v_C^2}{r} = mg \implies v_C^2 = gr \] --- ## Step 4: Substitute \( v_C^2 \) into Energy Equation From above: \[ v_^2 = gr + 2gh \] Given: - \( r = 2.40 \) ft - \( h = 7.5 \) ft - \( g = 32.2 \) ft/s² Calculate: \[ v_^2 = 32.2 \times 2.4 + 2 \times 32.2 \times 7.5 \] \[ v_^2 = 77.28 + 483 = 560.28 \] \[ v_ = \sqrt{560.28} = 23.691 \text{ ft/s} \] --- ## Step 5: Force Exerted by Package on Loop at C At C: \[ F = m \frac{v_C^2}{r} - mg \] But at the minimum speed (from above), \( m \frac{v_C^2}{r} = mg \), so the **force exerted by the package on the loop is \( \) lb** (it's just about to lose contact). --- ## **Final Answers** ### 1. The smallest velocity \( v_ \) for which the package will reach C: \[ \boxed{23.691 \ \text{ft/s}} \] ### 2. The force exerted by the package on the loop just before it leaves at C: \[ \boxed{.000 \ \text{lb}} \] --- ### **Summary Table** | Quantity | Value | |----------------------------------|---------------| | Smallest velocity \( v_ \) | 23.691 ft/s | | Force on loop at C | .000 lb |

# Solution to the Problem ## Problem Overview An 8 oz package is projected upward with an initial velocity \( v_0 \) by a spring, moving through a frictionless loop and reaching point C at height \( h = 7.5 \) ft. We need to determine: 1. The smallest velocity \( v_C \) for which the package will reach point C. 2. The force exerted by the package on the loop just before it leaves the loop at point C. ## Step-by-Step Solution ### Step 1: Convert Units - **Mass of the package**: \[ m = \frac{8 \text{ oz}}{16 \text{ oz/lb}} = 0.5 \text{ lb} \] - **Weight**: \[ W = mg = 0.5 \text{ lb} \quad (\text{where } g = 32.2 \text{ ft/s}^2) \] ### Step 2: Apply Conservation of Energy Considering the energy at points A (bottom) and C (top): - **At A**: - Kinetic Energy: \( \frac{1}{2} mv_0^2 \) - Potential Energy: 0 (reference level) - **At C**: - Kinetic Energy: \( \frac{1}{2} mv_C^2 \) - Potential Energy: \( mgh \) Setting total mechanical energy at A equal to total mechanical energy at C: \[ \frac{1}{2} mv_0^2 = \frac{1}{2} mv_C^2 + mgh \] \[ v_0^2 = v_C^2 + 2gh \] ### Step 3: Minimum Speed at Top (Point C) At the top of the loop (point C), for the package to just maintain contact, the centripetal force is provided by the weight: \[ m \frac{v_C^2}{r} = mg \] Thus, \[ v_C^2 = gr \] ### Step 4: Substitute \( v_C^2 \) into Energy Equation From the previous equation: \[ v_0^2 = gr + 2gh \] Given: - \( r = 2.40 \text{ ft} \) - \( h = 7.5 \text{ ft} \) - \( g = 32.2 \text{ ft/s}^2 \) Calculate: \[ v_0^2 = 32.2 \times 2.4 + 2 \times 32.2 \times 7.5 \] \[ v_0^2 = 77.28 + 483 = 560.28 \] \[ v_0 = \sqrt{560.28} \approx 23.691 \text{ ft/s} \] ### Step 5: Force Exerted by Package on Loop at C At point C, the net force exerted by the package on the loop is given by: \[ F = m \frac{v_C^2}{r} - mg \] When the package just maintains contact, \( m \frac{v_C^2}{r} = mg \), thus the normal force becomes zero: \[ F = 0 \text{ lb} \] ## Final Answers ### 1. The smallest velocity \( v_C \) for which the package will reach C: \[ \boxed{23.691 \text{ ft/s}} \] ### 2. The force exerted by the package on the loop just before it leaves at C: \[ \boxed{0.000 \text{ lb}} \] --- ### Summary Table | Quantity | Value | |----------------------------------|---------------| | Smallest velocity \( v_C \) | 23.691 ft/s | | Force on loop at C | 0.000 lb |