Let's break down the problem step by step: ### **Given:** - Mass of puck, \( m = 0.40 \) kg - Force \( F_1 = 4.0 \) N at \( \theta_1 = 20^\circ \) **below** the x-axis (so, -20° from x-axis) - Force \( F_2 = 7.0 \) N at \( \theta_2 = 50^\circ \) **above** the x-axis - **Frictionless** surface ### **Step 1: Decompose Forces into Components** #### **For \( F_1 \):** \[ \text{Components:} \] \[ F_{1x} = F_1 \cos(-20^\circ) = 4.0 \cos(-20^\circ) \] \[ F_{1y} = F_1 \sin(-20^\circ) = 4.0 \sin(-20^\circ) \] #### **For \( F_2 \):** \[ F_{2x} = F_2 \cos(50^\circ) = 7.0 \cos(50^\circ) \] \[ F_{2y} = F_2 \sin(50^\circ) = 7.0 \sin(50^\circ) \] --- ### **Step 2: Calculate Components Numerically** #### **\( F_1 \) Components:** - \( \cos(-20^\circ) = \cos(20^\circ) = 0.9397 \) - \( \sin(-20^\circ) = -0.3420 \) \[ F_{1x} = 4.0 \times 0.9397 = 3.7588 \text{ N} \] \[ F_{1y} = 4.0 \times (-0.3420) = -1.368 \text{ N} \] #### **\( F_2 \) Components:** - \( \cos(50^\circ) = 0.6428 \) - \( \sin(50^\circ) = 0.7660 \) \[ F_{2x} = 7.0 \times 0.6428 = 4.4996 \text{ N} \] \[ F_{2y} = 7.0 \times 0.7660 = 5.362 \text{ N} \] --- ### **Step 3: Add the Components** \[ F_{x,\text{total}} = F_{1x} + F_{2x} = 3.7588 + 4.4996 = 8.2584 \text{ N} \] \[ F_{y,\text{total}} = F_{1y} + F_{2y} = (-1.368) + 5.362 = 3.994 \text{ N} \] --- ### **Step 4: Find Magnitude of Net Force** \[ F_{\text{net}} = \sqrt{F_{x,\text{total}}^2 + F_{y,\text{total}}^2} \] \[ = \sqrt{(8.2584)^2 + (3.994)^2} \] \[ = \sqrt{68.187 + 15.952} \] \[ = \sqrt{84.139} \] \[ = 9.175 \text{ N} \] --- ### **Step 5: Find Direction of Net Force** \[ \theta = \arctan\left(\frac{F_{y,\text{total}}}{F_{x,\text{total}}}\right) = \arctan\left(\frac{3.994}{8.2584}\right) \] \[ = \arctan(0.484) \] \[ = 25.8^\circ \] **This angle is above the x-axis.** --- ### **Step 6: Find Acceleration** \[ a = \frac{F_{\text{net}}}{m} = \frac{9.175}{0.40} = 22.94 \text{ m/s}^2 \] --- ## **Final Answers** ### **Magnitude of acceleration:** \[ \boxed{22.9 \text{ m/s}^2} \] ### **Direction of acceleration:** \[ \boxed{25.8^\circ\ \text{above the x-axis}} \] --- If you need a diagram or further explanation, let me know!

The net acceleration of the puck is determined by vector addition of all applied forces, using their components along the coordinate axes. The resulting force's magnitude and direction are then obtained from the combined components, and the acceleration is found by dividing this net force by the mass.