# Step-by-Step Solution Given: - \( F_1 = 63.2 \) N (horizontal force) - \( F_2 = 17.1 \) N (applied at \( \theta = 30.^\circ \) **above the horizontal**) - Mass, \( m = 3.88 \) kg - Coefficient of kinetic friction, \( \mu_k = .200 \) - The block is moving to the right We are to find: - **(b) The acceleration \( a \) of the block in \(\mathrm{m/s^2}\) when \( \theta = 30.^\circ \)** --- ## Step 1: Draw a Free-Body Diagram Forces acting on the block: - Weight: \( mg \) (downward) - Normal force: \( F_N \) (upward) - \( F_1 \) (horizontal, right) - \( F_2 \) (at \( 30^\circ \) above the horizontal, right) - Kinetic friction: \( f_k \) (opposes motion, left) --- ## Step 2: Components of Forces **\( F_1 \):** - All horizontal: \( F_{1x} = F_1 = 63.2 \) N **\( F_2 \):** - Horizontal component: \( F_{2x} = F_2 \cos \theta = 17.1 \cos 30.^\circ \) - Vertical component: \( F_{2y} = F_2 \sin \theta = 17.1 \sin 30.^\circ \) Calculate: - \( \cos 30.^\circ = .866 \) - \( \sin 30.^\circ = .500 \) So: - \( F_{2x} = 17.1 \times .866 = 14.80 \) N - \( F_{2y} = 17.1 \times .500 = 8.55 \) N --- ## Step 3: Find the Normal Force (\( F_N \)) From vertical forces (given \( F_N = 46.60 \) N in part (a), but let's show): Upward: \( F_N \)\ Downward: \( mg \)\ Upward (from \( F_2 \)): \( F_{2y} \) Sum of vertical forces (no vertical acceleration): \[ F_N + F_{2y} - mg = \implies F_N = mg - F_{2y} \] Plug in values: \[ F_N = (3.88 \, \text{kg}) \times (9.80 \, \text{m/s}^2) - 8.55 \, \text{N} \] \[ F_N = 38.02 \, \text{N} - 8.55 \, \text{N} \] \[ F_N = 29.47 \, \text{N} \] But the given answer is \( 46.60 \) N, which suggests \( F_2 \) is **pushing down** (\( \theta = 30^\circ \) below horizontal). Let's check that case: \[ F_N = mg + F_{2y} \] \[ F_N = 38.02 \, \text{N} + 8.55 \, \text{N} \] \[ F_N = 46.57 \, \text{N} \] This matches the given answer. **So, \( F_2 \) is acting \( 30^\circ \) below the horizontal (pushing down).** --- ## Step 4: Find the Kinetic Friction Force (\( f_k \)) \[ f_k = \mu_k F_N = .200 \times 46.60 \, \text{N} = 9.32 \, \text{N} \] --- ## Step 5: Net Horizontal Force Sum of horizontal forces: \[ F_{\text{net},x} = F_1 + F_{2x} - f_k \] Calculate \( F_{2x} \): \[ F_{2x} = 17.1 \cos 30^\circ = 14.80 \, \text{N} \] So, \[ F_{\text{net},x} = 63.2 + 14.80 - 9.32 = 77.28 - 9.32 = 68. \, \text{N} \] --- ## Step 6: Calculate the Acceleration (\( a \)) \[ a = \frac{F_{\text{net},x}}{m} \] \[ a = \frac{68.68 \, \text{N}}{3.88 \, \text{kg}} \] \[ a = 17.7 \, \text{m/s}^2 \] --- ### Final Answer ## Acceleration of the Block \[ \boxed{a = 17.7 \ \text{m/s}^2} \] --- ### Summary Table | Quantity | Value | |--------------------|--------------| | Normal force (\( F_N \)) | 46.60 N | | Friction (\( f_k \)) | 9.32 N | | Net force (\( F_{\text{net},x} \)) | 68.68 N | | Acceleration (\( a \)) | 17.7 m/s\(^2\) |