Let's solve the problem step by step as requested. --- ## **Step 1: Find Initial Energy Stored in the Spring and Energy Needed to Reach Height \( h \)** **Given:** - Mass, \( m = .20\, \text{kg} \) - Spring constant, \( k = 2.5\, \text{N/cm} = 250\, \text{N/m} \) - Compression, \( x = 4.\, \text{cm} = .040\, \text{m} \) - Maximum height, \( h = 10\, \text{cm} = .10\, \text{m} \) **Initial energy stored in spring:** \[ E_{\text{spring}} = \frac{1}{2} k x^2 = \frac{1}{2} \cdot 250 \cdot (.040)^2 = .2\, \text{J} \] **Energy needed to reach height \( h \):** \[ E_{\text{grav}} = mgh = .20 \times 9.8 \times .10 = .196\, \text{J} \] **Explanation:** The spring stores \( .2\, \text{J} \) of energy. To reach height \( h \), the block needs \( .196\, \text{J} \). The difference must be the energy lost due to friction in region BC. --- ## **Step 2: Calculate Energy Lost to Friction and Find \(\mu\)** **Energy lost to friction:** \[ \Delta E = E_{\text{spring}} - E_{\text{grav}} = .2 - .196 = .004\, \text{J} \] **Work done by friction (as the block passes through BC twice):** - Friction force = \( f = \mu m g \) - Distance covered in BC = \( 2L \) (once on the way up, once on the way back) - \( L = 20\, \text{cm} = .20\, \text{m} \) \[ \text{Work by friction} = f \cdot d = \mu m g \cdot 2L \] Set equal to energy lost: \[ \mu m g \cdot 2L = .004 \] \[ \mu = \frac{.004}{2 \cdot .20 \cdot .20 \cdot 9.8} \] \[ \mu = \frac{.004}{.784} \] \[ \mu \approx .0051 \] **Explanation:** The total energy lost to friction (through BC twice) equals the difference between the spring energy and gravitational energy. Solving for \(\mu\) gives the coefficient of kinetic friction. --- ## **Final Answer** \[ \boxed{\mu \approx .0051} \] This is the coefficient of kinetic friction for the region between B and C.