Work As Area Under F-x Graphs: Videos & Practice Problems

To calculate work done on an object using a force versus displacement graph, determine the area under the curve between the graph and the x-axis. This involves breaking the graph into simpler shapes like rectangles and triangles. The area of a rectangle is calculated as base times height, while the area of a triangle is given by 12 × ⁢base⁢height. Positive areas indicate positive work, while negative areas indicate negative work, leading to the total work done being the sum of these areas.

Calculating Work As Area Under F-x Graphs

To calculate the work done on an object using a force versus displacement graph, one must understand that the work is represented by the area under the force (F) versus position (x) graph. This area is calculated between the graph and the x-axis, encompassing both positive and negative areas depending on the force's direction.

The fundamental principle is that the work done (W) can be determined by summing the areas of geometric shapes formed under the curve. For a graph that includes both rectangles and triangles, the areas can be calculated using basic geometry formulas:

1. **Area of a Rectangle**: The area (A) is calculated as:

\[ A = \text{base} \times \text{height} \]

2. **Area of a Triangle**: The area (A) is calculated as:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

When analyzing the graph, one can break it down into simpler shapes. For example, if the graph consists of one rectangle and two triangles, the areas can be calculated individually:

For the rectangle (Area 1), if the base is 4 units and the height is 30 units, the area is:

\[ \text{Area 1} = 4 \times 30 = 120 \text{ joules} \]

For the first triangle (Area 2), with a base of 12 units (from 4 to 16) and a height of 30 units, the area is:

\[ \text{Area 2} = \frac{1}{2} \times 12 \times 30 = 180 \text{ joules} \]

For the second triangle (Area 3), which is below the x-axis, the base is 4 units and the height is -10 units (indicating a downward force), the area is:

\[ \text{Area 3} = \frac{1}{2} \times 4 \times (-10) = -20 \text{ joules} \]

After calculating the areas, the total work done is found by summing these areas:

\[ W = \text{Area 1} + \text{Area 2} + \text{Area 3} = 120 + 180 - 20 = 280 \text{ joules} \]

This method simplifies the process of calculating work done by variable forces, allowing for a straightforward approach without needing to repeatedly apply the formula \( W = F \cdot d \cdot \cos(\theta) \). By visualizing the force as a function of position, one can efficiently determine the work done on an object.

Speed of a Remote Controlled Car Using F-x Graph

In this scenario, we analyze the motion of a remote control car subjected to a varying force over a distance. To determine the speed of the car at a specific point (x = 4), we utilize the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. The initial speed of the car is zero, meaning it starts from rest.

To find the final speed, we first calculate the work done on the car by determining the area under the force vs. distance graph. This area represents the work done in three segments: from 0 to 2, from 2 to 3, and from 3 to 4.

1. **Work from 0 to 2**: This area consists of a triangle and a rectangle. The triangle's area is calculated as:

Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 20 = 10 \, \text{joules} \)

The rectangle's area is:

Area of rectangle = \( \text{base} \times \text{height} = 1 \times 20 = 20 \, \text{joules} \)

Thus, the total work done from 0 to 2 is:

Work (0 to 2) = \( 10 + 20 = 30 \, \text{joules} \)

2. **Work from 2 to 3**: There is no force acting on the car in this segment, resulting in zero work done:

Work (2 to 3) = \( 0 \, \text{joules} \)

3. **Work from 3 to 4**: This area is a triangle with a base of 1 and a height of -15 (since it is below the x-axis). The area is calculated as:

Area of triangle = \( \frac{1}{2} \times 1 \times (-15) = -7.5 \, \text{joules} \)

Now, we sum the work done across all segments:

Total Work = Work (0 to 2) + Work (2 to 3) + Work (3 to 4) = \( 30 + 0 - 7.5 = 22.5 \, \text{joules} \)

According to the work-energy theorem, this total work is equal to the change in kinetic energy:

Net Work = \( K_{\text{final}} - K_{\text{initial}} \)

Since the initial kinetic energy is zero (the car starts from rest), we have:

22.5 joules = \( \frac{1}{2} m v_{\text{final}}^2 \)

Given that the mass of the car is 4 kg, we can substitute this value into the equation:

22.5 = \( \frac{1}{2} \times 4 \times v_{\text{final}}^2 \)

Solving for \( v_{\text{final}}^2 \):

22.5 = \( 2 v_{\text{final}}^2 \)

Thus, \( v_{\text{final}}^2 = \frac{22.5}{2} = 11.25 \)

Taking the square root gives:

v_{\text{final}} = \( \sqrt{11.25} \approx 3.35 \, \text{m/s} \)

Therefore, the speed of the car at x = 4 is approximately 3.35 meters per second.

Here’s what students ask on this topic:

How do you calculate work done using a force vs. displacement graph?

To calculate work done using a force vs. displacement graph, determine the area under the curve between the graph and the x-axis. This involves breaking the graph into simpler shapes like rectangles and triangles. The area of a rectangle is calculated as $\mathrm{base}\times \mathrm{height}$, while the area of a triangle is given by $1/2\times \mathrm{base}\times \mathrm{height}$. Positive areas indicate positive work, while negative areas indicate negative work. The total work done is the sum of these areas.

What does the area under a force vs. displacement graph represent?

The area under a force vs. displacement graph represents the work done on an object by the force. This area is calculated by summing the areas of simpler geometric shapes (rectangles and triangles) that lie between the graph and the x-axis. Positive areas correspond to positive work, while negative areas correspond to negative work. The total work is the sum of these individual areas.

How do you handle negative areas in a force vs. displacement graph?

Negative areas in a force vs. displacement graph occur when the force is below the x-axis, indicating that the force is acting in the opposite direction of the displacement. To handle these, calculate the area as usual but assign a negative sign to it. When summing the total work, include these negative areas to account for the direction of the force.

What is the formula for the area of a triangle in a force vs. displacement graph?

The formula for the area of a triangle in a force vs. displacement graph is $1/2\times \mathrm{base}\times \mathrm{height}$. Here, the base is the length along the displacement axis, and the height is the force value. This formula helps in calculating the work done by the force over the displacement.

Why is it easier to use a force vs. displacement graph for variable forces?

Using a force vs. displacement graph for variable forces is easier because it allows you to visually break down the problem into simpler geometric shapes. Instead of using complex calculus or averaging methods, you can calculate the work done by finding the area under the curve. This method simplifies the process, especially when dealing with non-constant forces.