A 0.171 kg screwdriver initially at rest, slides down a ramp that is inclined at \( 57.5^{\circ} \). What is the magnitude of the normal force of the screwdriver? \[ \mathrm{n}=[\text { ? ] N } \]

To find the magnitude of the normal force acting on the screwdriver, we start by noting that the normal force (\( n \)) is part of the forces at play when an object is on an inclined plane. The weight of the screwdriver can be calculated as \( mg = 0.171 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1.676 \, \text{N} \). The normal force is given by the equation: \[ n = mg \cos(\theta) \] Where \( \theta \) is the angle of the incline. So we have: \[ n = 1.676 \, \text{N} \times \cos(57.5^{\circ}) \] Calculating that gives: \[ n \approx 1.676 \, \text{N} \times 0.5446 \approx 0.912 \, \text{N} \] So the magnitude of the normal force is approximately \( 0.912 \, \text{N} \).