91Ó°ÊÓ

Work is a physical quantity that represents the amount of energy transferred or converted when a force is applied to an object, causing it to move along a certain distance. It is measured in Joules (J). Force is a vector quantity that represents the interaction between two objects, which can cause an acceleration, and is measured in Newtons (N). Displacement is a vector quantity that represents the change in an object's position and is measured in meters (m).

The dot product (also known as the scalar product or inner product) is an algebraic operation that takes two equal-length sequences (vectors) of numbers and returns a single number. In the context of the force and displacement vectors, the dot product can be used to determine the work done by multiplying the magnitudes of the vectors and the cosine of the angle between them. Mathematically, for two vectors A and B, the dot product is given by: \[ A \cdot B = ||A|| ||B|| \cos(\theta) \] Where \(||A||\) and \(||B||\) are the magnitudes of the vectors, and \(\theta\) is the angle between them.

To calculate the work done by a force in moving an object, we can use the formula: \[ W = F \cdot d = ||F|| ||d|| \cos(\theta) \] Where W is the work done, F is the force vector, d is the displacement vector, and \(\theta\) is the angle between the force and displacement vectors.

Suppose a force F = (3N, 4N) is applied to a point object which causes displacement d = (2m, 3m). Let's compute the work done by the force. 1. Calculate the magnitude of the force vector: \( ||F|| = \sqrt{(3^2 + 4^2)} = \sqrt{25} = 5 N \) 2. Calculate the magnitude of the displacement vector: \( ||d|| = \sqrt{(2^2 + 3^2)} = \sqrt{13} m \) 3. Determine the cosine of angle \(\theta\): Since \(F \cdot d = ||F|| ||d|| \cos(\theta)\), we can extract \( \cos(\theta) = \frac{F \cdot d}{||F|| ||d||}\) 4. Calculate the dot product of the force and displacement vectors: \( F \cdot d = (3 \times 2) + (4 \times 3) = 18 \) 5. Plug the values into the cosine formula: \( \cos(\theta) = \frac{18}{(5) (\sqrt{13})} \approx 0.948 \) 6. Calculate the work done: \( W = ||F|| ||d|| \cos(\theta) = (5) (\sqrt{13}) (0.948) \approx 15.485 J \) The work done by the force in moving the object is approximately 15.485 Joules.