A \( 50-\mathrm{N} \) crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of \( 37^{\circ} \) to the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker? 1) zero 2) 20 N 3) 30 N 4) 40 N 5) 50 N

To find the force applied by the worker, we first need to calculate the forces acting on the crate. The weight of the crate can be broken down into two components: the force acting parallel to the incline (which is due to gravity) and the frictional force. The gravitational force acting down the incline can be calculated using the formula: \[ F_{\text{gravity}} = mg \sin(\theta) \] Here, \( m \) is the mass (which we can find as \( \frac{50 \, \text{N}}{9.81 \, \text{m/s}^2} \), but for our purposes, we only need \( F_{\text{gravity}} \)), \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination \( 37^\circ \). The frictional force between the crate and the inclined plane is given as \( 10 \, \text{N} \). The net force acting against the worker’s force is the sum of the gravitational force down the incline and the frictional force. Thus, the worker needs to apply enough force to balance these out, expressed as: \[ F_{\text{applied}} = F_{\text{gravity}} + F_{\text{friction}} \] Calculating \( F_{\text{gravity}} = 50 \sin(37^\circ) \approx 30 \, \text{N} \) (using \( \sin(37^\circ) \approx 0.6 \)), we find that: \[ F_{\text{applied}} = 30 \, \text{N} + 10 \, \text{N} = 40 \, \text{N} \] Therefore, the force applied by the worker is **40 N**, so the correct answer is 4) 40 N.