Problem 1 - College Physics

Question

\(\cdot\) A fisherman reels in 12.0 \(\mathrm{m}\) of line while landing a fish, using a constant forward pull of 25.0 \(\mathrm{N}\) . How much work does the tension in the line do on the fish?

Step-by-Step Solution

Answer

300.0 Joules

Step 1: Identify the Given Values

We know the fisherman reels in 12.0 meters of line and the force applied is 25.0 Newtons. Let's denote the distance as \( d = 12.0 \, \text{m} \) and the force as \( F = 25.0 \, \text{N} \).

Step 2: Recall the Work Formula

The formula for work \( W \) is given by \( W = F \cdot d \cdot \cos(\theta) \), where \( \theta \) is the angle between the force and the direction of movement. In this problem, the force direction and the movement direction are the same, so \( \theta = 0 \) degrees.

Step 3: Calculate the Cosine Factor

Since \( \theta = 0 \) degrees, we have \( \cos(0) = 1 \). Therefore, the work done simplifies to \( W = F \cdot d \cdot 1 = F \cdot d \).

Step 4: Substitute the Values into the Formula

Now substitute the known values into the formula: \( W = 25.0 \, \text{N} \cdot 12.0 \, \text{m} \).

Step 5: Compute the Work Done

Multiply the force and distance to find the work done: \[ W = 25.0 \, \text{N} \times 12.0 \, \text{m} = 300.0 \, \text{J} \].

Step 6: Conclusion

The work done by the tension in the line on the fish is 300.0 Joules.

Key Concepts

Force

In physics, **force** is a fundamental concept that refers to any interaction that, when unopposed, will change the motion of an object. Newton defined force as \[ F = m imes a \] where

• \( F \) represents the force applied,
• \( m \) is the mass of the object, and
• \( a \) is the acceleration.

In simpler terms, force can be thought of as a push or pull acting upon an object resulting in movement or a change in an object's velocity.
The unit of force in the International System of Units (SI) is the Newton (N). When you hear someone say a force of 25 N was applied, it means that the force required to accelerate a 1 kg object by 1 meter per second squared was exerted.

Distance

In the physical sciences, **distance** is the measurement of how far apart two objects or points are. It plays a crucial role when calculating work done, as work is the product of force and the distance over which that force is applied.
In our exercise, the fisherman reeling in the fish moves the line a distance of 12.0 meters. This distance is measured in a straight line along the path of the movement. It is represented in physics by the symbol \( d \) and is often measured in meters (m).
Whenever you need to calculate work done, remember that it involves both the force exerted and the distance over which it acts. Without a clear understanding of the distance, you cannot find the accurate amount of work done.

Angle in Physics

An **angle in physics** often refers to the orientation between two lines, planes, or vectors. In work calculations, the angle is specifically crucial for understanding the direction of force relative to motion. The angle, represented by \( \theta \), decides how much of the force actually contributes to the movement.
For example, if you apply force perfectly in the direction of the movement, the angle is 0 degrees. This means every bit of force aids in moving. Conversely, a force applied at a 90-degree angle (like pushing a wall) does not move it forward, resulting in zero work done.
Knowing the angle helps in calculating the correct component of the applied force that accomplishes work, ensuring the calculation is both realistic and accurate.

Cosine Function

The **cosine function** is a mathematical tool used principally in trigonometry. It helps in work calculations by providing the ratio of the adjacent side to the hypotenuse in a right triangle. When dealing with work, the cosine function determines how much of the force is effectively used in performing work along the direction of movement.
For angles measured in relation to the straight path of motion, if the force directly aligns with it, \( \cos(0) = 1 \), meaning the entire force contributes to the work. Conversely, \( \cos(90^{\circ}) = 0 \), so no contribution to work occurs for perpendicular forces.
Thus, cosine is essential in physics for discerning force effectiveness. It refines work calculations by factoring in the direction of the force relative to an object's movement.