Of course! Let's break down and solve the problem step by step in English. --- ### **Problem Summary** - A spring with constant \( k = 200 \) N/m is attached to a 2 kg cylinder. - The cylinder slides freely on a horizontal guide. - At \( t = 0 \), a constant force of 10 N is applied to the cylinder. - Initially, the spring is **not deformed** (not stretched or compressed), and the system is **at rest**. - Find: 1. The velocity of the cylinder when \( x = 40 \) mm. 2. The **maximum displacement** of the cylinder. --- ## **Step 1: List the Given Data** - Spring constant: \( k = 200 \) N/m - Mass: \( m = 2 \) kg - Applied force: \( F = 10 \) N - Initial position: \( x_0 = 0 \) - Initial speed: \( v_0 = 0 \) - Position to check: \( x = 40 \) mm \( = 0.04 \) m --- ## **Step 2: Energy Approach (Work-Energy Principle)** ### **Work done by the force:** The work done by the constant force as the cylinder moves from 0 to \( x \): \[ W = F \cdot x \] ### **Spring Energy Stored:** The energy stored in the spring when stretched by \( x \): \[ U = \frac{1}{2} k x^2 \] ### **Kinetic Energy:** The kinetic energy when the cylinder has velocity \( v \): \[ K = \frac{1}{2} m v^2 \] ### **Work-Energy Equation:** Initial energy = 0 (at rest, spring not deformed) \[ \text{Work by force} = \text{Increase in kinetic energy} + \text{Increase in spring potential energy} \] \[ F x = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \] --- ## **Step 3: Find the Velocity at \( x = 40 \) mm** Plug in the values: - \( F = 10 \) N - \( x = 0.04 \) m - \( m = 2 \) kg - \( k = 200 \) N/m \[ 10 \cdot 0.04 = \frac{1}{2} \cdot 2 \cdot v^2 + \frac{1}{2} \cdot 200 \cdot (0.04)^2 \] \[ 0.4 = 1 \cdot v^2 + 100 \cdot 0.0016 \] \[ 0.4 = v^2 + 0.16 \] \[ v^2 = 0.4 - 0.16 = 0.24 \] \[ v = \sqrt{0.24} \approx 0.49 \text{ m/s} \] --- ## **Step 4: Find the Maximum Displacement** At maximum displacement, the velocity is zero (\( v = 0 \)) because the object momentarily stops before coming back. Use the work-energy equation with \( v = 0 \) and let \( x = x_{max} \): \[ F x_{max} = \frac{1}{2} k x_{max}^2 \] \[ 10 x_{max} = 100 x_{max}^2 \] \[ 100 x_{max}^2 - 10 x_{max} = 0 \] \[ x_{max}(100 x_{max} - 10) = 0 \] \[ x_{max} = 0 \quad \text{or} \quad 100 x_{max} = 10 \] \[ x_{max} = \frac{10}{100} = 0.1 \text{ m} = 100 \text{ mm} \] --- ## **Summary of Answers** 1. **Velocity at \( x = 40 \) mm:** \[ v \approx 0.49 \text{ m/s} \] 2. **Maximum displacement:** \[ x_{max} = 0.1 \text{ m} = 100 \text{ mm} \] --- If you need a deeper explanation or a step-by-step for any part, just let me know!