Understanding and Calculating Work Done by a Force in Physics

Work is a fundamental concept in physics, representing the energy transferred to or from an object via the application of force. Calculating work involves understanding the relationship between force, displacement, and the angle between them. This article will walk through a detailed explanation of how to calculate work done by a force acting in the direction of a given vector, using examples and mathematical illustrations.

Calculating Work Done by a Force

To calculate the work done by a force when an object is moved from one point to another, we use the formula:

W vec{F} cdot vec{d}

where vec{F} is the force vector and vec{d} is the displacement vector. Let's break down the process using a specific example where a 10 N force acts in the direction of the vector 1 1.

Step-by-Step Calculation

Calculate the Displacement Vector vec{d}

The displacement from point P -21 to point Q56 can be calculated as follows:

vec{d} Q - P 5 - (-2) 6 - (-1) 7 5

Normalize the Direction of the Force

The force of 10 N acts in the direction of the vector 1 1. The first step is to normalize this vector to find the unit vector.

text{Magnitude of } 1 1 sqrt{1^2 1^2} sqrt{2}

The unit vector in the direction of 1 1 is:

hat{u} left left frac{1}{sqrt{2}} frac{1}{sqrt{2}} right right

Thus, the force vector vec{F} can be calculated as:

vec{F} 10 hat{u} 10 left left frac{1}{sqrt{2}} frac{1}{sqrt{2}} right right left left frac{10}{sqrt{2}} frac{10}{sqrt{2}} right right 5sqrt{2} 5sqrt{2}

Calculate the Work Done

The work done can now be calculated using the dot product:

vec{F} cdot vec{d} (5sqrt{2} 5sqrt{2}) cdot (7 5) 5sqrt{2} cdot 7 5sqrt{2} cdot 5 35sqrt{2} 25sqrt{2} 60sqrt{2}

Therefore, the work done is:

W 60sqrt{2} text{ Joules}

For a numerical approximation, sqrt{2} approx 1.414, so:

W approx 60 times 1.414 approx 84.84 text{ Joules}

Thus, the work done is approximately 84.84 Joules or exactly 60sqrt{2} Joules.

Alternative Calculation Method

Another approach to calculating work involves breaking down the force into its horizontal and vertical components and using the respective displacements. For a horizontal force of 10 cos 45° and a displacement of 7 units, the work done is:

7 times 10 cos 45° 7 times 10 times frac{1}{sqrt{2}} 70 times frac{1}{1.414} approx 49.5 text{ work units}

Similarly, for a vertical force of 10 sin 45° and a vertical displacement of 5 units, the work done is:

5 times 10 sin 45° 5 times 10 times frac{1}{sqrt{2}} 50 times frac{1}{1.414} approx 35.36 text{ work units}

Therefore, the total work done is:

49.5 35.36 84.86 text{ work units}

Illustrating the Motion on a Coordinate System

On a coordinate system, the motion and the force can be visualized as follows. Draw the motion along the x-axis and the force vector at an angle to the motion. Extend the lines to form a triangle. The angle between the force and the motion can be determined. The work can then be calculated using the formula:

Work is force times distance times the cosine of the angle between force and the direction of the motion.

W F.D where F is the force vector, and D is the distance vector.

Here, F 10(1/sqrt{2}, 1/sqrt{2}) and D (7, 5). Therefore, the work done is:

Work done 5sqrt{2}7 51 60sqrt{2} Nm

Assuming P and Q are specified in meter units.