A 2000 kg loaded coal cart is pulled up at an angle \( 10^{\circ} \) inclined at a onstant speed by the cable parallel to the incline. The frictional forc mpeding the motion is 800 N . How much work is done on the cart by he cable as the cart is pulled 30 m . Use \( g=9.81 \mathrm{~ms} \). A 2 kg object is dropped from the height of 12 m and is moving at ms \( \square \) just before hitting the ground. How large is the average rictional force retarding the motion. Use \( g=9.81 \mathrm{~m} / \mathrm{s} 2 \)

Let's address both parts of your question step by step. --- ### **1. Work Done by the Cable on the Coal Cart** **Given:** - Mass of the cart, \( m = 2000 \, \text{kg} \) - Angle of incline, \( \theta = 10^\circ \) - Frictional force opposing motion, \( F_{\text{friction}} = 800 \, \text{N} \) - Distance pulled, \( d = 30 \, \text{m} \) - Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) **Steps:** 1. **Calculate the Component of Gravitational Force Along the Incline:** \[ F_{\text{gravity, parallel}} = m \cdot g \cdot \sin(\theta) \] \[ F_{\text{gravity, parallel}} = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \sin(10^\circ) \] \[ \sin(10^\circ) \approx 0.1736 \] \[ F_{\text{gravity, parallel}} \approx 2000 \times 9.81 \times 0.1736 \approx 3402 \, \text{N} \] 2. **Determine the Total Force Exerted by the Cable:** Since the cart is moving at a constant speed, the cable must counteract both the friction and the component of gravity. \[ F_{\text{cable}} = F_{\text{friction}} + F_{\text{gravity, parallel}} = 800 \, \text{N} + 3402 \, \text{N} = 4202 \, \text{N} \] 3. **Calculate the Work Done by the Cable:** \[ W = F_{\text{cable}} \times d = 4202 \, \text{N} \times 30 \, \text{m} = 126,060 \, \text{J} \] **Rounded:** \[ W \approx 1.26 \times 10^5 \, \text{J} \] --- ### **2. Average Frictional Force Retarding a Falling Object** **Given:** - Mass of the object, \( m = 2 \, \text{kg} \) - Height, \( h = 12 \, \text{m} \) - Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) **Assumption:** The object is moving at a **constant speed** just before hitting the ground. This implies that the net force acting on it is zero, meaning the frictional force exactly balances the gravitational force. **Steps:** 1. **Determine the Gravitational Force:** \[ F_{\text{gravity}} = m \cdot g = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \] 2. **Since the Object Moves at Constant Speed:** The average frictional force, \( F_{\text{friction}} \), must balance the gravitational force. \[ F_{\text{friction}} = F_{\text{gravity}} = 19.62 \, \text{N} \] --- ### **Final Answers:** 1. **Work done by the cable:** \( \mathbf{1.26 \times 10^5 \, \text{J}} \) 2. **Average frictional force:** \( \mathbf{19.62 \, \text{N}} \)