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? \[ n=[?] \mathrm{N} \]

To find the magnitude of the normal force of the screwdriver, we need to consider the forces acting on the screwdriver as it slides down the ramp. Given: - Mass of the screwdriver, \( m = 0.171 \, \text{kg} \) - Angle of inclination of the ramp, \( \theta = 57.5^{\circ} \) The forces acting on the screwdriver are: 1. Weight of the screwdriver, \( W = mg \) 2. Normal force, \( n \) 3. Frictional force, \( f \) Since the screwdriver is initially at rest, the net force acting on it is zero. Therefore, the sum of the forces in the horizontal direction is zero, and the sum of the forces in the vertical direction is also zero. Let's denote the horizontal component of the weight as \( W_{\text{horizontal}} \) and the vertical component of the weight as \( W_{\text{vertical}} \). The horizontal component of the weight is given by: \[ W_{\text{horizontal}} = W \cos(\theta) \] The vertical component of the weight is given by: \[ W_{\text{vertical}} = W \sin(\theta) \] Since the screwdriver is at rest, the net force in the horizontal direction is zero, which means the normal force is equal to the horizontal component of the weight: \[ n = W_{\text{horizontal}} = W \cos(\theta) \] Now, we can calculate the magnitude of the normal force using the given values. The magnitude of the normal force of the screwdriver is approximately 0.97396 N.