A student pushes an \( M=15.6 \) kg shopping cart across the room by applying a \( \left|\vec{F}_{A}\right|=185 \mathrm{~N} \) force at an angle of \( \theta=37.0^{\circ} \) below the horizontal as shown in the figure. ( \( \mu=0.430 \) ) Find the magnitude of the weight \( \left|\vec{F}_{g}\right| \) of the cart. \( \left|\vec{F}_{g}\right|= \) Find the magnitude of the normal force \( \left|\vec{F}_{N}\right| \) on the cart. \( \left|\vec{F}_{N}\right|= \) Find the magnitude of the force of friction \( \left|\vec{F}_{f}\right| \) on the cart. \( \left|\vec{F}_{f}\right|=\square \) Find the magnitude of the acceleration \( |\vec{a}| \) of the cart. \( |\vec{a}|=\square \)

We are given:  Mass of cart, M = 15.6 kg  Applied force, |Fₐ| = 185 N (directed 37.0° below the horizontal)  Coefficient of kinetic friction, μ = 0.430  Acceleration due to gravity, g ≈ 9.80 m/s² (you may also use 9.81 m/s²) We wish to find: 1. The weight |F_g| of the cart 2. The normal force |F_N| on the cart 3. The frictional force |F_f| 4. The acceleration |a| of the cart ───────────────────────────── Step 1. Find the weight |F_g|: The weight is given by   |F_g| = M·g Plug in the numbers:   |F_g| = (15.6 kg)(9.80 m/s²) ≈ 152.88 N Rounded suitably, we have   |F_g| ≈ 153 N ───────────────────────────── Step 2. Find the normal force |F_N|: Because the applied force is directed downward at an angle below the horizontal, its vertical component adds to the weight. (Assuming the cart is not accelerating vertically, the sum of forces in the vertical direction must vanish.) Resolve the applied force into components:  Horizontal: Fₐₓ = |Fₐ| cosθ  Vertical (downward): Fₐ_y = |Fₐ| sinθ Since both weight and Fₐ_y act downward, the normal force from the floor must counter both. Thus,   N = |F_g| + |Fₐ| sinθ Now, using θ = 37.0°, note that:   sin 37.0° ≈ 0.6018 Then:   |Fₐ| sinθ = (185 N)(0.6018) ≈ 111.33 N Now compute N:   N = 152.88 N + 111.33 N ≈ 264.21 N Rounded,   |F_N| ≈ 264 N ───────────────────────────── Step 3. Find the frictional force |F_f|: Friction is given by   |F_f| = μ·N Plug in:   |F_f| = 0.430 × 264.21 N ≈ 113.66 N Rounded,   |F_f| ≈ 114 N ───────────────────────────── Step 4. Find the acceleration |a| of the cart: The net horizontal force is given by the horizontal component of the applied force minus the friction opposing the motion. First, resolve the applied force horizontally:   Fₐₓ = |Fₐ| cosθ Using cos 37.0° ≈ 0.7986, we have   Fₐₓ = (185 N)(0.7986) ≈ 147.68 N Now, net horizontal force:   F_net = Fₐₓ – |F_f|   F_net ≈ 147.68 N – 113.66 N ≈ 34.02 N Use Newton’s second law in the horizontal direction:   F_net = M·a ⟹ a = F_net / M Thus,   a = (34.02 N)/(15.6 kg) ≈ 2.18 m/s² ───────────────────────────── Final Answers: 1. The weight of the cart:  |F_g| ≈ 153 N 2. The normal force on the cart: |F_N| ≈ 264 N 3. The frictional force:    |F_f| ≈ 114 N 4. The acceleration of the cart: |a| ≈ 2.18 m/s² These results complete the solution.