A \(52.3-\mathrm{kg}\) trunk is pushed \(5.95 \mathrm{~m}\) at constant speed up a \(28.0^{\circ}\) incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is \(0.19 .\) Calculate the work done by ( \(a\) ) the applied force and \((b)\) the force of gravity.

The force applied on the trunk is the force required to overcome the force of gravity and the frictional force, since the trunk is moving at constant speed. This means that the sum of forces acting on the trunk is zero. Let \( F_a \) denote the applied force, \( F_g \) the force of gravity and \( F_f \) the frictional force. So we have: \( F_a - (F_g + F_f) = 0 \). This leads to: \( F_a = F_g + F_f \). The trunk has a mass of \( 52.3 kg \), the angle of incline is \( 28.0^\circ \) and the coefficient of kinetic friction is \( 0.19 \). \n\nWe can find \( F_g \) and \( F_f \) as follows: For \( F_g \), it will be the component of the weight \( mg \) which is acting perpendicular to the incline, that will be \( mg \sin \theta \). So, \( F_g = 52.3kg \times 9.8m/s^2 \times \sin(28.0^\circ) \). \n\nAs for \( F_f \), it is given by the product of the normal force and the friction coefficient, hence \( F_f = μN \), where \( μ \) is the coefficient of friction and \( N \) is the normal force, acting perpendicular to the plane. Here, \( N = mg \cos \theta \), so \( F_f = μ mg \cos \theta = 0.19 \times 52.3kg \times 9.8m/s^2 \times \cos(28.0^\circ) \). \n\nSubstituting those values into the equation \( F_a = F_g + F_f \) will yield the value for the applied force \( F_a \).

The work done by a force is given by the formula \( W = Fd \cos \theta \), where \( W \) is the work done, \( F \) is the force, \( d \) is the distance over which the force is exerted and \( \theta \) is the angle between the force and the direction of motion. Here, \( F \) is the applied force \( F_a \), \( d \) is the distance the trunk is pushed up the incline \( 5.95m \), and \( θ \) is the angle of inclination \( 28.0^\circ \). As the applied force is horizontal and the displacement of the trunk is along the incline, the angle between them is \( 90^\circ - 28.0^\circ \). Hence, substituting these values into the work formula will yield the work done by the applied force.

Now, we examine the work done by gravity. The force due to gravity is \( F_g = mg \sin \theta \), which we calculated in Step 1. Since gravity acts downward and the trunk is pushed upward along the incline, the angle between the force of gravity and the direction of motion is \( 180^\circ - 28.0^\circ \). Hence, substituting these values into the work formula will yield the work done by the force of gravity.