Understanding Work Done by Gravity

The question asks for the work done by the force of gravity on a $40$ kg object as it moves horizontally from point A to point B on a table.

Physics Principles Involved

Work done ($W$) by a constant force ($\vec{F}$) causing a displacement ($\vec{d}$) is defined as:

$W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos(\theta)$

Where:

• $|\vec{F}|$ is the magnitude of the force.
• $|\vec{d}|$ is the magnitude of the displacement.
• $\theta$ is the angle between the force vector and the displacement vector.

Analyzing the Forces and Displacement

In this problem:

• Force: The force we are considering is gravity. The force of gravity ($ \vec{F}_g $) always acts vertically downwards, towards the center of the Earth.
• Displacement: The object is moved horizontally from point A to point B. This means the displacement vector ($\vec{d}$) is along the horizontal direction.

Calculating the Angle

Since the gravitational force acts vertically downwards and the object's movement is purely horizontal, the angle ($\theta$) between the gravitational force vector and the displacement vector is $90^\circ$.

$ \theta = 90^\circ $

Calculating Work Done by Gravity

Now, we can substitute the values into the work formula:

$ W_{\text{gravity}} = |\vec{F}_g| |\vec{d}| \cos(90^\circ) $

We know that the cosine of $90^\circ$ is $0$:

$ \cos(90^\circ) = 0 $

Therefore, the work done by gravity is:

$ W_{\text{gravity}} = |\vec{F}_g| |\vec{d}| \times 0 $

$ W_{\text{gravity}} = 0 \text{ J} $

Conclusion

The mass of the object ($40$ kg) and the distance it moved horizontally do not affect the work done by gravity in this specific scenario because the gravitational force is perpendicular to the direction of motion. When the force is perpendicular to the displacement, the work done by that force is always zero.