Work Done Calculation Vectorially

In physics, work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. When dealing with vectors, calculating work done involves considering both the magnitude and direction of the force and displacement vectors.

Understanding Vectors

Before diving into the calculation of work done vectorially, it's important to understand what vectors are. A vector is a quantity that has both magnitude and direction. In the context of work, the two vectors we are concerned with are:

• Force (F): A vector that represents the push or pull on an object.
• Displacement (s): A vector that represents the change in position of an object.

Scalar Product (Dot Product)

The calculation of work done vectorially involves the scalar product, also known as the dot product, of the force and displacement vectors. The scalar product of two vectors A and B is given by:

[ \textbf{A} \cdot \textbf{B} = |\textbf{A}| |\textbf{B}| \cos(\theta) ]

where ( |\textbf{A}| ) and ( |\textbf{B}| ) are the magnitudes of vectors A and B, respectively, and ( \theta ) is the angle between them.

Work Done Formula

The work done ( W ) by a constant force ( \textbf{F} ) acting on an object as it moves through a displacement ( \textbf{s} ) is given by the dot product of the force and displacement vectors:

[ W = \textbf{F} \cdot \textbf{s} ] [ W = |\textbf{F}| |\textbf{s}| \cos(\theta) ]

where ( \theta ) is the angle between the force and displacement vectors.

Important Points and Differences

Aspect	Description
Direction of Vectors	Work is only done in the direction of the force.
Angle ( \theta )	The angle between the force and displacement vectors is crucial in calculating work done.
Zero Work	If ( \theta ) is 90 degrees, no work is done since ( \cos(90^\circ) = 0 ).
Negative Work	If ( \theta ) is greater than 90 degrees, work is negative, indicating work done against the force.
Units	Work is measured in Joules (J) in the SI system.

Examples

Example 1: Work Done by a Horizontal Force

A force of 10 N is applied horizontally to move a box 5 meters to the right. Calculate the work done.

Solution:

Here, the force and displacement are in the same direction, so ( \theta = 0^\circ ).

[ W = |\textbf{F}| |\textbf{s}| \cos(\theta) ] [ W = (10 \, \text{N}) (5 \, \text{m}) \cos(0^\circ) ] [ W = 50 \, \text{J} ]

Example 2: Work Done by a Force at an Angle

A force of 10 N is applied at an angle of 30 degrees to the horizontal to move a box 5 meters to the right. Calculate the work done.

Solution:

Here, the force is at an angle to the displacement, so ( \theta = 30^\circ ).

[ W = |\textbf{F}| |\textbf{s}| \cos(\theta) ] [ W = (10 \, \text{N}) (5 \, \text{m}) \cos(30^\circ) ] [ W = 50 \, \text{J} \times \frac{\sqrt{3}}{2} ] [ W \approx 43.3 \, \text{J} ]

Example 3: Work Done Against Gravity

A force is applied to lift a 10 kg object vertically upwards by 5 meters. Calculate the work done against gravity.

Solution:

The force required to lift the object is equal to its weight, which is the product of mass and gravitational acceleration (g = 9.8 m/s²).

[ |\textbf{F}| = m \times g ] [ |\textbf{F}| = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 ] [ |\textbf{F}| = 98 \, \text{N} ]

Since the force is in the same direction as the displacement (upwards), ( \theta = 0^\circ ).

[ W = |\textbf{F}| |\textbf{s}| \cos(\theta) ] [ W = (98 \, \text{N}) (5 \, \text{m}) \cos(0^\circ) ] [ W = 490 \, \text{J} ]

In summary, calculating work done vectorially requires an understanding of vectors, the scalar product, and the relationship between force, displacement, and the angle between them. The examples provided illustrate how to apply the formula for work done in different scenarios.