Work Done By A Force Pushing A 1 Kg Block Answering Physics Question

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Hey guys! Let's dive into a classic physics problem that involves calculating the work done by a force. This is a fundamental concept in mechanics, and understanding it is crucial for tackling more complex problems. So, let's break down the problem step-by-step and make sure we grasp the key ideas. Our focus will be on understanding work done, kinetic energy, and how they relate to each other.

Problem Statement

Imagine a 1 kg block sitting at rest. We then apply a force, F, to this block, pushing it along a smooth, frictionless surface. As the block passes a certain point, let's call it point B, it's moving at a speed of 6 m/s. The big question we need to answer is this: How much work did the force F do on the block to get it moving at that speed? We will explore the relationship between force and work done in a practical scenario.

The problem provides us with a few possible answers, and our goal is to figure out which one is correct. The options are:

  • A) 18 J
  • B) 20 J
  • C) 28 J
  • D) 30 J
  • E) 36 J

So, let's get started and figure out how to calculate the work done in this scenario!

Understanding the Concepts

Before we jump into the calculations, let's quickly review some key concepts. This will help us approach the problem with a solid understanding of the physics involved. We're primarily dealing with work and energy, specifically kinetic energy. Make sure you understand the definitions and formulas of kinetic energy and work done.

Work

In physics, work is done when a force causes an object to move a certain distance. The amount of work done depends on the magnitude of the force, the distance over which it acts, and the angle between the force and the direction of motion. When the force is in the same direction as the displacement, the work done is simply the product of the force and the distance. However, in this problem, we're going to use a different approach that relies on the work-energy theorem, which makes things a bit simpler. The work-energy theorem simplifies the calculation by relating work to the change in kinetic energy.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It depends on the object's mass and its velocity. A heavier object moving at the same speed will have more kinetic energy than a lighter one. Similarly, an object moving faster will have more kinetic energy than the same object moving slower. The formula for kinetic energy (KE) is:

KE = 1/2 * m * v^2

where m is the mass and v is the velocity. Understanding kinetic energy is fundamental to solving this problem efficiently. We'll see how it helps in calculating the work done directly.

The Work-Energy Theorem

The work-energy theorem is a crucial link between work and energy. It states that the work done on an object is equal to the change in its kinetic energy. In other words, if you do work on an object, you're changing its speed, and that change in speed directly translates to a change in kinetic energy. Mathematically, this is expressed as:

W = ΔKE = KE_final - KE_initial

where W is the work done, ΔKE is the change in kinetic energy, KE_final is the final kinetic energy, and KE_initial is the initial kinetic energy. Using the work-energy theorem simplifies our task immensely, as we don't need to know the exact force or the distance over which it acted. The work-energy theorem is a shortcut that makes the solution straightforward.

Applying the Work-Energy Theorem to Solve the Problem

Now that we've got our concepts straight, let's apply the work-energy theorem to solve our problem. This theorem is our key to finding the work done by the force. We'll start by calculating the initial and final kinetic energies of the block.

1. Calculate the Initial Kinetic Energy

Initially, the block is at rest, which means its velocity is 0 m/s. Using the formula for kinetic energy, we can calculate the initial kinetic energy (KE_initial):

KE_initial = 1/2 * m * v_initial^2

Plugging in the values, we get:

KE_initial = 1/2 * 1 kg * (0 m/s)^2 = 0 J

So, the initial kinetic energy of the block is 0 Joules since it started from rest. Starting from rest simplifies the initial energy calculation.

2. Calculate the Final Kinetic Energy

At point B, the block has a velocity of 6 m/s. We can now calculate the final kinetic energy (KE_final) using the same formula:

KE_final = 1/2 * m * v_final^2

Substituting the given values:

KE_final = 1/2 * 1 kg * (6 m/s)^2 = 1/2 * 1 kg * 36 m2/s2 = 18 J

Therefore, the final kinetic energy of the block at point B is 18 Joules. The final kinetic energy is crucial for applying the work-energy theorem.

3. Apply the Work-Energy Theorem

Now we can use the work-energy theorem to find the work done by the force F. The theorem states:

W = ΔKE = KE_final - KE_initial

We already calculated KE_initial and KE_final, so we can plug those values in:

W = 18 J - 0 J = 18 J

So, the work done by the force F is 18 Joules. This straightforward calculation confirms our final answer.

The Answer

Based on our calculations, the work done by the force F on the 1 kg block is 18 Joules. Looking back at the options provided, we can see that:

  • A) 18 J
  • B) 20 J
  • C) 28 J
  • D) 30 J
  • E) 36 J

The correct answer is A) 18 J. The calculation using the work-energy theorem directly gives us this result.

Key Takeaways

Let's recap the important points we've learned from this problem:

  • Work-Energy Theorem: The work done on an object is equal to the change in its kinetic energy. This theorem simplifies the calculation of work when we know the initial and final velocities. The work-energy theorem is a powerful tool in mechanics.
  • Kinetic Energy: The kinetic energy of an object is given by KE = 1/2 * m * v^2. Understanding this formula is essential for solving problems involving motion and energy. Knowing how to calculate kinetic energy is vital.
  • Problem-Solving Approach: By breaking down the problem into steps (calculating initial KE, calculating final KE, and applying the work-energy theorem), we can solve it systematically. A structured approach makes complex problems manageable.

Additional Tips and Tricks

To further enhance your problem-solving skills, consider these additional tips:

  • Units: Always pay attention to the units. In this problem, we used kilograms (kg) for mass, meters per second (m/s) for velocity, and Joules (J) for work and energy. Consistent units are crucial for accurate calculations.
  • Frictionless Surfaces: The problem stated that the surface was smooth, meaning we could ignore friction. If friction were present, the work done would be different, as some energy would be lost as heat due to friction. Friction introduces additional complexity to the problem.
  • Visualize the Problem: Try to visualize the scenario. This can help you understand the forces acting on the object and how they affect its motion. Visualization is a powerful tool for problem-solving.

Practice Problems

To solidify your understanding, try solving similar problems. Here's a variation you can attempt:

  • A 2 kg block is pushed by a force such that when passing through point B, it has a speed of 8 m/s. Determine the amount of work done by the force. Try changing the mass and velocity to create new problems.

By practicing different variations, you'll become more confident in applying the work-energy theorem and solving similar physics problems. Regular practice is key to mastering these concepts.

Conclusion

So there you have it! We've successfully calculated the work done by a force on a moving block using the work-energy theorem. This problem illustrates the fundamental relationship between work and energy, and how understanding these concepts can help us solve real-world physics problems. Keep practicing, and you'll become a pro at these types of questions in no time! Remember, mastering the work-energy theorem is a significant step in understanding physics.