Question

To push a \(25-\mathrm{kg}\) crate up a \(27^{\circ}\) incline, a worker exerts a force of \(120 \mathrm{~N}\), parallel to the incline. As the crate slides \(3.6 \mathrm{~m}\), how much work is done on the crate by \((a)\) the worker, \((b)\) the force of gravity, and \((c)\) the normal force of the incline?

Short answer

The work done by the worker is \(432 \mathrm{~J}\), by the force of gravity is \(-675 \mathrm{~J}\) and by the normal force of the incline is \(0 \mathrm{~J}\).

Step-by-step solution

Calculation of work done by the worker

The work done by the worker is calculated using the formula: \(Work = Force \times Distance\). Here, the force is given as \(120 \mathrm{~N}\) and the distance is \(3.6 \mathrm{~m}\), thus, \(Work_{worker} = 120 \mathrm{~N} \times 3.6 \mathrm{~m} = 432 \mathrm{~J}\).

Calculation of work done by the force of gravity

The work done by gravity is calculated using the formula: \(Work = Force \times Distance \times cos(\Theta)\). Here, the force is the weight of the crate, which is \(25 \mathrm{kg} \times 9.8 m/s^2 = 245 \mathrm{~N}\). The angle of incline is \(27^\circ\), therefore, \(Work_{gravity} = 245 \mathrm{~N} \times 3.6 \mathrm{~m} \times cos(27^\circ) = -675 \mathrm{~J}\). The negative sign indicates that the work done by gravity is in the opposite direction to the motion of the crate.

Calculation of work done by the normal force

The normal force acts perpendicular to the direction of motion, and the work done by a force perpendicular to the direction of motion is always zero. Thus, \(Work_{normal} = 0\).

Summarize the results and make conclusions

The work done by the worker on the crate is \(432 \mathrm{~J}\), by the gravity is \(-675 \mathrm{~J}\) and by the normal force is \(0 \mathrm{~J}\). The negative sign in the work done by gravity indicates that this work opposes the motion of the crate.

Key concepts

Inclined Plane

The inclined plane, in this exercise, is a flat surface tilted at an angle. It's a common tool used to minimize the effort needed to lift objects. By spreading the height over a longer distance, the same weight can be moved with less force than lifting it vertically.

In a typical inclined plane problem, like this one, a force is applied parallel to the surface, and we observe how various factors play their roles. Here, the inclined plane is angled at 27 degrees, which significantly impacts the forces involved. The force applied is not required to completely lift the object but rather to move it along the plane.

Understanding the relationship between the angle and forces acting is crucial. It helps determine how much force is needed and calculates the effects of gravity when an object moves on a slope.

Work Done by a Force

Work done by a force is a core concept in physics, and it's calculated as the product of force and displacement in the direction of the force. Mathematically, it's expressed as:

\[ Work = Force \times Distance \times \cos(\Theta) \]

In this problem, the worker exerts a force of 120 N parallel to the plane and moves the crate a distance of 3.6 m up the incline. The work done by the worker is straightforward as the force's direction aligns with the crate's movement. This results in a positive amount of work, calculated as 432 Joules.

Understanding how work works in directions parallel or not parallel to an object's motion is vital for analyzing energy transfers.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object resting on it. On an inclined plane, the normal force acts perpendicular to the inclined surface.

It's crucial to note that the normal force doesn't do any work on the object because it doesn't cause displacement in its direction. This is why in our problem, where the crate slides along the surface of the incline, the work done by the normal force is zero.

The normal force balances part of the gravitational force, keeping the object from sinking into the surface. Calculating this force can be a precursor to understanding friction and stability, but as for work done, it plays no direct role.

Force of Gravity

The force of gravity pulls objects towards the earth and is always directed downward. For the crate on the incline, gravity acts not in line with the movement but at an angle.

To compute the work done by gravity, we use the component of the gravitational force parallel to the incline. In mathematical terms:

\[Work_{gravity} = weight \times distance \times \cos(\Theta)\]

The weight of the crate results in a gravitational force of 245 N. However, due to the 27 degrees angle, the actual work done by gravity comes out to -675 Joules. The negative sign indicates that this work opposes the direction of movement, as gravity tries to pull the crate back down the incline.

Exploring gravity in this context shows how it affects objects on different planes and the importance of angles in such scenarios.