67. A constant unbalanced force of friction acts on a 15.0 -kilogram mass moving along a horizontal surface at 10.0 meters per second. If the mass is breught to rest in 1.50 seconds, what is the magnitude of the force of friction? \( \begin{array}{ll}\text { A) } 10.0 \mathrm{~N} & \text { B) } 100 \mathrm{~N} \\ \text { C) } 147 \mathrm{~N} & \text { D) } 150 . \mathrm{N}\end{array} \)

To find the force of friction, we can start by calculating the acceleration of the mass as it comes to rest. Using the formula \( a = \frac{Δv}{Δt} \), where \( Δv = 0 - 10.0 \, m/s = -10.0 \, m/s \) and \( Δt = 1.50 \, s \), we get \( a = \frac{-10.0 \, m/s}{1.50 \, s} \approx -6.67 \, m/s^2 \). Now, using Newton’s second law \( F = m \cdot a \), where \( m = 15.0 \, kg \), the force of friction \( F = 15.0 \, kg \cdot (-6.67 \, m/s^2) \). This results in \( F \approx -100.05 \, N \). Since we care about the magnitude of the force, it is approximately \( 100 \, N \). Therefore, the answer is **B) 100 N**. In historical physics, the concept of friction was famously discussed by Sir Isaac Newton, who established the laws of motion that explain how frictional forces act on bodies in motion. The insight into these forces paved the way for advancements in mechanics and engineering, influencing everything from car design to sports equipment. In real-world scenarios, understanding friction is crucial for various applications, such as ensuring that tires have enough grip on the road or creating effective braking systems. Engineers routinely calculate frictional forces when designing vehicles or machinery to prevent accidents and ensure efficiency, demonstrating that these principles permeate our daily lives.