QUESTION 3 FRICTION 3.1 State the difference between static friction and kinetic friction. (2) 2 A block with a weight of 100 N is placed on an incline plane 4 at is \( 20^{\circ} \) to the horizontal. The coefficient of friction between the block and the surface is 0,24 Calculate: 2.1 The weight component parallel to the plane. 2.1 The weight component perpendicular to the plane. 3 The frictional force. 4 The smallest force required to pull the body up the plane. STION 4 HEAT The amount of heat energy absorbed or given off by a substance depends on THREE factors. Name these THREE factors. Calculate the efficiency with which 8 kg of coal is completely burnt in order to produce 90 MJ of heat energy. kg of steel is cooled down in oil and the temperature drops from \( 900^{\circ} \mathrm{C} \) to \( 7^{\circ} \mathrm{C} \). The specific heat capacity of steel is \( 0,46 \mathrm{~kJ} / \mathrm{kg}^{\circ} \mathrm{C} \) and that of oil is \( 5 \mathrm{~kJ} / \mathrm{kg}{ }^{\circ} \mathrm{C} \). Calculate the quantity of heat lost by the steel N 5 HYDRAULICS ine the densitv of a substance.

Let's break down the problem step by step. ### 3.1 Difference between Static Friction and Kinetic Friction 1. **Static Friction**: This is the frictional force that prevents two surfaces from sliding past each other. It acts when an object is at rest and is trying to move. The force of static friction increases with the applied force until it reaches a maximum value, beyond which the object will start to move. 2. **Kinetic Friction**: This is the frictional force acting between two surfaces that are sliding past each other. It is generally less than the maximum static friction and remains constant regardless of the speed of the sliding object. ### 2.1 Weight Component Parallel to the Plane To find the weight component parallel to the incline, we can use the formula: \[ F_{\parallel} = W \cdot \sin(\theta) \] where: - \( W = 100 \, \text{N} \) (weight of the block) - \( \theta = 20^{\circ} \) ### 2.2 Weight Component Perpendicular to the Plane To find the weight component perpendicular to the incline, we can use the formula: \[ F_{\perpendicular} = W \cdot \cos(\theta) \] ### 3. Frictional Force The frictional force can be calculated using the formula: \[ F_{\text{friction}} = \mu \cdot F_{\perpendicular} \] where: - \( \mu = 0.24 \) (coefficient of friction) ### 4. Smallest Force Required to Pull the Body Up the Plane The smallest force required to pull the body up the plane can be calculated using: \[ F_{\text{required}} = F_{\parallel} + F_{\text{friction}} \] Now, let's perform the calculations for each part. #### Calculations 1. Calculate \( F_{\parallel} \) 2. Calculate \( F_{\perpendicular} \) 3. Calculate \( F_{\text{friction}} \) 4. Calculate \( F_{\text{required}} \) Let's compute these values. Calculate the value by following steps: - step0: Calculate: \(100\sin\left(20\right)\) Calculate or simplify the expression \( 100 * \cos(20) \). Calculate the value by following steps: - step0: Calculate: \(100\cos\left(20\right)\) Calculate or simplify the expression \( (100 * \sin(20)) + (0.24 * (100 * \cos(20))) \). Calculate the value by following steps: - step0: Calculate: \(\left(100\sin\left(20\right)\right)+\left(0.24\left(100\cos\left(20\right)\right)\right)\) - step1: Remove the parentheses: \(\left(100\sin\left(20\right)\right)+\left(0.24\times 100\cos\left(20\right)\right)\) - step2: Multiply the terms: \(100\sin\left(20\right)+\left(0.24\times 100\cos\left(20\right)\right)\) - step3: Multiply the terms: \(100\sin\left(20\right)+24\cos\left(20\right)\) - step4: Calculate the trigonometric value: \(101.088495\) Calculate or simplify the expression \( 0.24 * (100 * \cos(20)) \). Calculate the value by following steps: - step0: Calculate: \(0.24\left(100\cos\left(20\right)\right)\) - step1: Remove the parentheses: \(0.24\times 100\cos\left(20\right)\) - step2: Multiply the terms: \(24\cos\left(20\right)\) ### Calculated Values 1. **Weight Component Parallel to the Plane**: \[ F_{\parallel} = 100 \cdot \sin(20^{\circ}) \approx 34.19 \, \text{N} \] 2. **Weight Component Perpendicular to the Plane**: \[ F_{\perpendicular} = 100 \cdot \cos(20^{\circ}) \approx 93.69 \, \text{N} \] 3. **Frictional Force**: \[ F_{\text{friction}} = 0.24 \cdot F_{\perpendicular} \approx 0.24 \cdot 93.69 \approx 22.49 \, \text{N} \] 4. **Smallest Force Required to Pull the Body Up the Plane**: \[ F_{\text{required}} = F_{\parallel} + F_{\text{friction}} \approx 34.19 + 22.49 \approx 56.68 \, \text{N} \] ### Summary of Results - Weight component parallel to the plane: \( \approx 34.19 \, \text{N} \) - Weight component perpendicular to the plane: \( \approx 93.69 \, \text{N} \) - Frictional force: \( \approx 22.49 \, \text{N} \) - Smallest force required to pull the body up the plane: \( \approx 56.68 \, \text{N} \) ### 4. Factors Affecting Heat Energy The amount of heat energy absorbed or given off by a substance depends on: 1. The mass of the substance. 2. The specific heat capacity of the substance. 3. The change in temperature of the substance. ### Efficiency Calculation To calculate the efficiency of burning 8 kg of coal to produce 90 MJ of heat energy, we can use the formula: \[ \text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Energy Input}} \right) \times 100 \] Assuming the energy input is based on the energy content of the coal, we need to know the energy content of the coal to proceed with this calculation. ### Heat Lost by Steel To calculate the quantity of heat lost by the steel, we can use the formula: \[ Q = m \cdot c \cdot \Delta T \] where: - \( m \) = mass of steel (unknown) - \( c = 0.46 \, \text{kJ/kg°C} \) (specific heat capacity of steel) - \( \Delta T = 900 - 7 = 893 \, \text{°C} \) Please provide the mass of the steel to calculate the heat lost.