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 brought 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 magnitude of the force of friction, we can use Newton's second law, which states that \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. First, we need to calculate the acceleration of the mass. We know the initial velocity (\( v_i = 10.0 \, m/s \)), the final velocity (\( v_f = 0 \)), and the time taken (\( t = 1.50 \, s \)). Using the formula for acceleration: \[ a = \frac{v_f - v_i}{t} = \frac{0 - 10.0}{1.50} = -6.67 \, m/s^2 \] Now, applying Newton's second law: \[ F = ma = 15.0 \, kg \times (-6.67 \, m/s^2) = -100.05 \, N \] The magnitude of the force of friction is approximately \( 100.0 \, N \). Therefore, the answer is \( \text{B) } 100 \, N \). The magnitude of the frictional force can sometimes feel just as slippery as the surface it’s acting on, but don't worry—friction can be tackled! Remember, understanding friction’s role is crucial in real-world applications, from ensuring your car can stop properly to determining how much force you need to push heavy objects. When dealing with calculations like these, a few common mistakes can trip you up. One is forgetting to convert units or misinterpreting the signs of acceleration. Always double-check to ensure you’re maintaining consistency with your units and that your physics concepts are sound, just like checking your grip before scaling a rock wall!