Let's break down the problem step by step: ### **Problem Statement** A force \( F \) is applied to a 100 kg block as shown in the figure (at an angle θ = 37° to the horizontal). The block accelerates at \( a = 2 \, \text{m/s}^2 \). The coefficient of kinetic friction between the block and the plane is \( \mu_k = 0.25 \). You are to **calculate the magnitude of force \( F \)**. --- ### **Solution Steps** #### 1. **List Known Quantities** - Mass, \( m = 100 \) kg - Acceleration, \( a = 2 \) m/s² - Coefficient of kinetic friction, \( \mu_k = 0.25 \) - Angle, \( \theta = 37^\circ \) - \( g = 9.8 \) m/s² #### 2. **Draw a Free Body Diagram** - \( F \) (applied force, at angle θ above horizontal) - \( f_k \) (kinetic friction, opposite direction) - \( N \) (normal reaction, upward) - \( mg \) (weight, downward) #### 3. **Resolve Forces** - **Horizontal (x-direction):** \[ F_x = F \cos \theta \] - **Vertical (y-direction):** \[ F_y = F \sin \theta \] #### 4. **Normal Force (\( N \)):** \[ N + F \sin \theta = mg \] \[ N = mg - F \sin \theta \] #### 5. **Friction Force (\( f_k \)):** \[ f_k = \mu_k N = \mu_k (mg - F \sin \theta) \] #### 6. **Net Force in x-direction:** \[ F \cos \theta - f_k = ma \] \[ F \cos \theta - \mu_k (mg - F \sin \theta) = ma \] \[ F \cos \theta - \mu_k mg + \mu_k F \sin \theta = ma \] \[ F (\cos \theta + \mu_k \sin \theta) = ma + \mu_k mg \] \[ F = \frac{ma + \mu_k mg}{\cos \theta + \mu_k \sin \theta} \] --- ### **Plug in the values** - \( m = 100 \) - \( a = 2 \) - \( \mu_k = 0.25 \) - \( g = 9.8 \) - \( \theta = 37^\circ \) - \( \cos 37^\circ \approx 0.7986 \) - \( \sin 37^\circ \approx 0.6018 \) \[ F = \frac{(100 \times 2) + 0.25 \times 100 \times 9.8}{0.7986 + 0.25 \times 0.6018} \] Calculate numerator: - \( 100 \times 2 = 200 \) - \( 0.25 \times 100 \times 9.8 = 245 \) - Numerator: \( 200 + 245 = 445 \) Calculate denominator: - \( 0.25 \times 0.6018 = 0.15045 \) - \( 0.7986 + 0.15045 = 0.94905 \) Now, \[ F = \frac{445}{0.94905} \approx 469 \text{ N} \] --- ### **Closest Answer** - The correct option is **480 N** (since it's the closest available answer). --- ### **Final Answer:** **480 N**

Based on the detailed calculation, the magnitude of the applied force \( F \) is approximately **469 N**, which is closest to **540 N** among the options provided. However, since 540 N is not listed, and the options are 80, 540, 601, 682, the best match is **540 N**. **Correct answer: 540 N**