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. ### Question 3: Friction #### 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. 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. Block on an Incline Plane Given: - Weight of the block, \( W = 100 \, \text{N} \) - Angle of incline, \( \theta = 20^{\circ} \) - Coefficient of friction, \( \mu = 0.24 \) #### 2.1 Weight Component Parallel to the Plane The weight component parallel to the incline can be calculated using the formula: \[ W_{\parallel} = W \cdot \sin(\theta) \] #### 2.2 Weight Component Perpendicular to the Plane The weight component perpendicular to the incline can be calculated using the formula: \[ W_{\perpendicular} = W \cdot \cos(\theta) \] #### 3. Frictional Force The frictional force can be calculated using the formula: \[ F_{\text{friction}} = \mu \cdot W_{\perpendicular} \] #### 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 the formula: \[ F_{\text{required}} = W_{\parallel} + F_{\text{friction}} \] Now, let's perform the calculations for each part. ### Calculations 1. Calculate \( W_{\parallel} \) 2. Calculate \( W_{\perpendicular} \) 3. Calculate \( F_{\text{friction}} \) 4. Calculate \( F_{\text{required}} \) Let's proceed with these calculations. Calculate the value by following steps: - step0: Calculate: \(100\cos\left(20\right)\) 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: \(24\cos\left(20\right)\) Calculate or simplify the expression \( 100 * \sin(20) \). Calculate the value by following steps: - step0: Calculate: \(100\sin\left(20\right)\) 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: \(101.088495\) ### Results of Calculations 1. **Weight Component Parallel to the Plane**: \[ W_{\parallel} = 100 \cdot \sin(20^{\circ}) \approx 34.19 \, \text{N} \] 2. **Weight Component Perpendicular to the Plane**: \[ W_{\perpendicular} = 100 \cdot \cos(20^{\circ}) \approx 93.69 \, \text{N} \] 3. **Frictional Force**: \[ F_{\text{friction}} = \mu \cdot W_{\perpendicular} = 0.24 \cdot 93.69 \approx 22.49 \, \text{N} \] 4. **Smallest Force Required to Pull the Body Up the Plane**: \[ F_{\text{required}} = W_{\parallel} + F_{\text{friction}} \approx 34.19 + 22.49 \approx 56.68 \, \text{N} \] ### 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: \( \approx 56.68 \, \text{N} \) ### Question 4: Heat The amount of heat energy absorbed or given off by a substance depends on three factors: 1. **Mass of the substance**: More mass means more heat energy is required to change its temperature. 2. **Specific heat capacity**: Different substances require different amounts of heat to change their temperature. 3. **Change in temperature**: The greater the temperature change, the more heat energy is absorbed or released. #### Efficiency Calculation Given: - Mass of coal = 8 kg - Heat energy produced = 90 MJ = \( 90 \times 10^6 \, \text{J} \) To calculate the efficiency, we need to know the energy content of the coal. Assuming the energy content of coal is approximately \( 24 \, \text{MJ/kg} \): \[ \text{Total energy from coal} = 8 \, \text{kg} \times 24 \, \text{MJ/kg} = 192 \, \text{MJ} \] Efficiency can be calculated using the formula: \[ \text{Efficiency} = \left( \frac{\text{Useful energy output}}{\text{Total energy input}} \right) \times 100 \] Now, let's calculate the efficiency. Calculate the value by following steps: - step0: Calculate: \(\left(\frac{90}{192}\right)\times 100\) - step1: Reduce the fraction: \(\frac{15}{32}\times 100\) - step2: Reduce the numbers: \(\frac{15}{8}\times 25\) - step3: Multiply: \(\frac{15\times 25}{8}\) - step4: Multiply: \(\frac{375}{8}\) ### Efficiency Calculation Using the formula for efficiency: \[ \text{Efficiency} = \left( \frac{90 \, \text{MJ}}{192 \, \text{MJ}} \right) \times 100 \approx 46.88\% \] ### Summary of Efficiency - The efficiency with which 8 kg of coal is completely burnt to produce 90 MJ of heat energy is approximately **46.88%**. ### Heat Loss Calculation Given: - Mass of steel = \( m \) kg (not provided, so we will denote it as \( m \)) - Initial temperature of steel = \( 900^{\circ}C \) - Final temperature of steel = \( 7^{\circ}C \) - Specific heat capacity of steel = \( 0.46 \, \text{kJ/kg}^{\circ}C \) The quantity of heat lost by the steel can be calculated using the formula: \[ Q = m \cdot c \cdot \Delta T \] where: - \( Q \) = heat lost - \( c \) = specific heat capacity of steel - \( \Delta T = T_{\text{initial}} - T_{\text{final}} = 900 - 7 = 893^{\circ}C \) Now, let's express the heat lost by the steel in terms of \( m \). \[ Q = m \cdot 0.46 \cdot 893 \] If you provide the mass of the steel, I can calculate the exact quantity of heat lost. Otherwise, the expression for heat lost is: \[ Q = 410.78m \, \text{kJ} \] ### Question 5: Hydraulics The density of a substance is defined as its mass per unit volume. It can be expressed mathematically as: \[ \rho = \frac{m}{V} \] where: - \( \rho \) = density - \( m \) = mass - \( V \) = volume If you have any further questions or need additional calculations, feel free to ask!