Question

To push a \(52-\mathrm{kg}\) crate across a floor, a worker applies a force of \(190 \mathrm{~N}\), directed \(22^{\circ}\) below the horizontal. As the crate moves \(3.3 \mathrm{~m}\), how much work is done on the crate by \((a)\) the worker, (b) the force of gravity, and \((c)\) the normal force of the floor on the crate?

Short answer

The work done on the crate by the worker is \(W_{\text{worker}} = (190 \cos 22^{\circ}) \times 3.3\) J, the work done by the force of gravity is \(W_{\text{gravity}} = 0\) J, and the work done by the normal force is \(W_{\text{normal}} = 0\) J.

Step-by-step solution

Work done by the worker

The force applied by the worker is at an angle, so only the horizontal component of the force actually does the work. This can be calculated as \(F_{\text{worker},x} = F_{\text{worker}} \cos \theta = 190 \cos 22^{\circ}\). The work then is \(W_{\text{worker}} = F_{\text{worker},x} \times d = (190 \cos 22^{\circ}) \times 3.3\)

Work done by gravity

The force of gravity is acting downwards, but the displacement is horizontal. Since the angle between the directions of the force of gravity and the displacement is 90 degrees, and \(\cos 90^{\circ} = 0\), the work done by gravity is zero. So, \(W_{\text{gravity}} = 0\).

Work done by the normal force

The normal force is also acting perpendicular to the direction of movement. Like the gravitational force, it doesn't do any work because its direction is 90 degrees to the displacement. So, \(W_{\text{normal}} = 0\).

Key concepts

Work-Energy Principle

The work-energy principle is a fundamental concept in physics that relates the work done on an object to the change in its kinetic energy. When a force is applied to an object causing it to move, work is done. According to the work-energy principle, the total work done by all forces acting on an object is equal to the change in its kinetic energy.

In mathematical terms, this principle can be expressed as: \[ W = \Delta KE \] where \( W \) is the work done, and \( \Delta KE \) is the change in kinetic energy. If the object has a significant acceleration or deceleration, then the kinetic energy changes, and work is being done. It’s important to understand that if the total work is zero, the kinetic energy remains constant, indicating that the object will continue to move at the same speed, or remain stationary if initially at rest. In the problem provided, the work done by the worker leads to the movement of the crate, indicating that the energy is transferred from the worker to the kinetic energy of the crate.

The work-energy principle is incredibly useful for solving physics problems as it directly links forces and motion, providing a clear framework for understanding how different forces affect the energy and movement of an object.

Force Components

Understanding force components is crucial when analyzing the work done by a force in any given direction. Forces can have both magnitude and direction, making them vector quantities. When a force is not aligned with the direction of motion, it's helpful to break it down into components that are parallel and perpendicular to the direction of movement.

For example, the force applied by the worker in our exercise problem has both horizontal and vertical components due to its angle of application. This leads to the concept of force components, where a single force can be resolved into two components:

• \( F_x = F \cos(\theta) \) - the horizontal component.
• \( F_y = F \sin(\theta) \) - the vertical component.

The horizontal component (\( F_x \)) is responsible for the work done in moving the object along the floor, whereas the vertical component (\( F_y \)) relates to forces like lifting against gravity or pushing down against a surface, which in this case doesn't contribute to the work in the direction of the crate's movement. By calculating these components, we can accurately determine the work done in any direction of interest.

Angle of Force Application

The angle at which a force is applied significantly influences the amount of work done on an object. This is because the effectiveness of a force in doing work is dependent on how much of the force is acting in the direction of the object's displacement.

In the context of physics problems, when calculating work done, the angle \( \theta \) between the force and the direction of displacement must be considered. The formula for work done is given by:\[ W = Fd \cos(\theta) \]Where:\( W \) is work done,
\( F \) is the magnitude of the force,
\( d \) is the displacement of the object, and
\( \theta \) is the angle between the force and the displacement vector.If \( \theta = 0^\circ \), the force is fully effective in doing work since it's completely aligned with the displacement. Conversely, if \( \theta = 90^\circ \), no work is done because the force is completely perpendicular to the direction of displacement. In problems like our exercise, where the force is applied at a non-zero and non-right angle, only a portion of the force contributes to the work, highlighting the importance of calculating the correct component of the force that lines up with the displacement. This concept is vital for correctly applying the equation for work and understanding the physical implications of forces applied at various angles.