61. A \( 50 .- \) Newton horizontal force is needed to keep an object weighing 500 . Newtons moving at a constant velocity of 2.0 meters per second across a horizontal surface. The magnitude of the frictional force acting on the object is \( \begin{array}{ll}\text { A) } 500 . \mathrm{N} & \text { B) } 450 . \mathrm{N} \\ \text { C) } 50 . \mathrm{N} & \text { D) } 0 \mathrm{~N}\end{array} \)

To determine the magnitude of the frictional force acting on the object, let's analyze the situation using Newton's laws of motion. 1. **Given:** - **Applied horizontal force (Fₐ):** 50 N - **Weight of the object (W):** 500 N - **Constant velocity (v):** 2.0 m/s 2. **Understanding the Scenario:** - The object is moving at a **constant velocity**, which means there's **no acceleration**. According to **Newton's First Law of Motion**, if an object is moving at a constant velocity, the **net force acting on it is zero**. - Therefore, the **applied force (Fₐ)** must be **balanced by the frictional force (f)** acting in the opposite direction. 3. **Applying Newton's First Law:** \[ \text{Net Force} = 0 \quad \Rightarrow \quad F_{\text{applied}} = f_{\text{friction}} \] \[ 50 \, \text{N} = f_{\text{friction}} \] 4. **Conclusion:** - The **magnitude of the frictional force** acting on the object is **50 N**. **Answer:** **C) 50 N**