Let’s break down the problem step by step, focusing on both **(a)** (force of friction) and **(b)** (velocity as it enters the trash can). --- ### **Given Data** - Mass of **dinner bowl** (\(m_b\)) = 0.250 kg - Mass of **kibble** (\(m_k\)) = 0.0150 kg - Initial speed of dinner bowl (\(v_{b0}\)) = 7.32 m/s - Angle (\(\theta\)) = 30.0º north of due east - Distance to trash (\(d\)) = 20.0 m east - Coefficient of kinetic friction (\(\mu_k\)) = 0.500 - Dinner bowl is launched horizontally from ground level --- ## **A. Force of friction on the dinner cannon** The frictional force (\(f_k\)) is given by: \[ f_k = \mu_k N \] where \(N\) is the normal force. On a horizontal surface, \(N = mg\). So, \[ f_k = \mu_k (m_b + m_k)g \] \[ = 0.500 \times (0.250 + 0.0150) \times 9.8 \] \[ = 0.500 \times 0.265 \times 9.8 \] \[ = 0.500 \times 2.597 \] \[ = 1.2985\, \text{N} \] **Final answer (a):** \[ \boxed{1.30\, \text{N}} \] (rounded to 3 significant figures) --- ## **B. Velocity (magnitude and direction) of dinner and kibble as it enters the trash can** Since friction is ignored (as per instructions), only the initial velocity and direction matter. ### **Step 1: Decompose Initial Velocity** The initial velocity (\(v_{b0}\)) is at 30º north of east. - \(v_{bx} = v_{b0} \cos(30^\circ) = 7.32 \times 0.866 = 6.34\, \text{m/s}\) (east) - \(v_{by} = v_{b0} \sin(30^\circ) = 7.32 \times 0.5 = 3.66\, \text{m/s}\) (north) ### **Step 2: Relative Motion of Kibble** The kibble is shot north from the bowl with velocity \(v_{k/b} = 10.0\, \text{m/s}\). So, total velocity of kibble relative to ground: - \(v_{kx} = v_{bx} = 6.34\, \text{m/s}\) (east) - \(v_{ky} = v_{by} + 10.0 = 3.66 + 10.0 = 13.66\, \text{m/s}\) (north) ### **Step 3: Magnitude and Direction** \[ v_k = \sqrt{v_{kx}^2 + v_{ky}^2} = \sqrt{(6.34)^2 + (13.66)^2} \] \[ = \sqrt{40.1956 + 186.5956} \] \[ = \sqrt{226.7912} \] \[ = 15.06\, \text{m/s} \] Angle north of east: \[ \theta = \tan^{-1}\left(\frac{v_{ky}}{v_{kx}}\right) = \tan^{-1}\left(\frac{13.66}{6.34}\right) \] \[ = \tan^{-1}(2.155) \] \[ \approx 65.2^\circ \] --- ### **Final Answer (b):** \[ \boxed{ \text{Velocity magnitude: } 15.1\, \text{m/s}, \quad \text{Direction: } 65.2^\circ \text{ north of east} } \] --- **Summary:** - (a) Friction force: **1.30 N** - (b) Velocity as it enters the trash: **15.1 m/s at 65.2º north of east**

Frictional force opposes motion and is calculated as the product of the coefficient of kinetic friction and the normal force, which equals the weight of the object on a horizontal surface. The velocity of a projectile combines its horizontal and vertical components, determined by initial launch angle and speed, to find its magnitude and direction at any point.