Calculating Work Done By A Force Physics Problem Solved

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In physics, work is defined as the energy transferred to or from an object by the application of force along with a displacement. It's a scalar quantity, meaning it has magnitude but no direction. Understanding how to calculate work is crucial in various fields, from mechanics to engineering. This article aims to provide a comprehensive explanation of work done by a force, focusing on a specific problem involving a force applied at an angle to the displacement.

The Formula for Work

The work (W) done by a constant force (F) on an object that undergoes a displacement (d) is given by the formula:

W = F d cos(θ)

Where:

  • W is the work done (measured in joules, J)
  • F is the magnitude of the force (measured in newtons, N)
  • d is the magnitude of the displacement (measured in meters, m)
  • θ is the angle between the force vector and the displacement vector (measured in degrees)

This formula highlights several important aspects of work:

  • Force and Displacement: Work is done only when a force causes a displacement. If there is no displacement (d = 0), then no work is done (W = 0), regardless of the magnitude of the force.
  • Angle Matters: The angle (θ) between the force and displacement is crucial. The cosine of the angle determines how much of the force contributes to the work. When the force is in the same direction as the displacement (θ = 0°), cos(0°) = 1, and the work done is maximum (W = F d). When the force is perpendicular to the displacement (θ = 90°), cos(90°) = 0, and no work is done (W = 0). When the force opposes the displacement (θ = 180°), cos(180°) = -1, and the work done is negative (W = -F d), indicating that the force is taking energy away from the object.
  • Scalar Quantity: Work is a scalar quantity, meaning it has magnitude but no direction. The sign of the work (positive or negative) indicates whether the force is adding energy to the object (positive work) or taking energy away from the object (negative work).

Problem Statement

Let's consider the problem presented: What is the value of work done on an object when a 0.1 x 10²-newton force moves it 30 meters and the angle between the force and the displacement is 25°?

This problem requires us to apply the work formula to calculate the work done in a specific scenario. We are given the following information:

  • Force (F) = 0.1 x 10² N = 10 N
  • Displacement (d) = 30 m
  • Angle (θ) = 25°

Step-by-Step Solution

To solve this problem, we will follow these steps:

  1. Identify the given values: We have already identified the force (F), displacement (d), and angle (θ).
  2. Write down the formula for work: W = F d cos(θ)
  3. Substitute the given values into the formula: W = (10 N) * (30 m) * cos(25°)
  4. Calculate the cosine of the angle: Using a calculator, we find that cos(25°) ≈ 0.906
  5. Multiply the values to find the work: W = (10 N) * (30 m) * (0.906) W = 300 Nâ‹…m * 0.906 W = 271.8 J

Therefore, the work done on the object is approximately 271.8 joules.

Analyzing the Answer Choices

The given answer choices are:

  • A. 27 x 10² joules
  • B. 3.0 x 10² joules
  • C. 7.5 x 10² joules
  • D. 0 joules

Comparing our calculated value (271.8 J) with the answer choices, we can see that it is closest to A. 27 x 10² joules (which is 2700 J). However, our calculated value is significantly smaller. Let's re-examine the problem statement and our calculations to ensure accuracy.

Upon review, we notice an error in the interpretation of answer choice A. 27 x 10² joules is equal to 2700 Joules, and answer choice B which is 3.0 x 10² joules is 300 Joules. 271.8 is closest to 300. So, the correct answer should be B. 3.0 x 10² joules.

Key Concepts and Considerations

  • Units: It is crucial to use consistent units in the work formula. The standard units are newtons (N) for force, meters (m) for displacement, and joules (J) for work. One joule is equal to one newton-meter (1 J = 1 Nâ‹…m).
  • Positive and Negative Work: Positive work indicates that the force is adding energy to the object, causing it to speed up or increase its kinetic energy. Negative work indicates that the force is taking energy away from the object, causing it to slow down or decrease its kinetic energy. For example, the work done by friction is typically negative because friction opposes the motion and reduces the object's kinetic energy.
  • Net Work: When multiple forces act on an object, the net work done is the sum of the work done by each individual force. The net work is equal to the change in the object's kinetic energy (Work-Energy Theorem).
  • Variable Forces: The formula W = F d cos(θ) applies when the force is constant and the displacement is along a straight line. If the force is variable or the path is curved, we need to use integration to calculate the work done.

Common Mistakes to Avoid

  • Incorrect Angle: Ensure that the angle used in the formula is the angle between the force vector and the displacement vector. Sometimes, the problem might provide a different angle, and you need to calculate the correct angle.
  • Unit Conversions: Always use consistent units. If the force is given in kilonewtons (kN) and the displacement is in centimeters (cm), you need to convert them to newtons (N) and meters (m) before applying the formula.
  • Sign of Work: Remember that work can be positive, negative, or zero. The sign depends on the angle between the force and the displacement.
  • Misinterpreting Scientific Notation: Pay close attention to scientific notation when interpreting answer choices and performing calculations.

Real-World Applications

The concept of work is fundamental in many areas of physics and engineering. Here are a few examples:

  • Mechanical Engineering: Calculating the work done by engines, motors, and other mechanical systems is essential for designing efficient machines.
  • Civil Engineering: Understanding work is crucial for designing structures that can withstand forces and stresses, such as bridges and buildings.
  • Sports: Analyzing the work done by athletes in various sports, such as lifting weights or running, helps optimize training techniques.
  • Everyday Life: We encounter work in our daily lives whenever we push, pull, or lift objects. For example, the work done in lifting a grocery bag or pushing a lawnmower can be calculated using the work formula.

Practice Problems

To solidify your understanding of work, try solving the following practice problems:

  1. A 50-N force is applied to a box, causing it to move 5 meters horizontally. If the force is applied at an angle of 30° to the horizontal, how much work is done?
  2. A crane lifts a 1000-kg car vertically 15 meters. How much work is done by the crane?
  3. A person pushes a stalled car with a force of 200 N, but the car does not move. How much work is done?

Conclusion

Calculating the work done by a force is a fundamental concept in physics. By understanding the work formula and its implications, we can analyze and solve a wide range of problems in mechanics, engineering, and everyday life. Remember to pay attention to the angle between the force and displacement, use consistent units, and consider the sign of the work. By practicing and applying these concepts, you can master the calculation of work and its applications. In the given problem, we correctly calculated the work done to be 271.8 Joules, which is closest to answer choice B. 3.0 x 10² joules after correcting our initial interpretation of the options.

This article provides a comprehensive guide to understanding and calculating work done by a force, ensuring a strong foundation in this essential physics concept.