Work Done By Force F1: Calculation & Explanation

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Hey guys! Today, we're diving into a physics problem where we need to figure out the work done by a force, specifically force F1, on a block that starts off chilling at rest. We're given that the block travels a distance of 8 meters, and our mission is to calculate just how much work F1 puts in to make that happen. Let's break it down step by step so you can totally nail this type of problem.

Understanding the Basics of Work

Before we jump into the calculations, let's quickly recap what "work" actually means in physics. In simple terms, 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 the object moves, and the angle between the force and the direction of motion. The formula for work is:

W = F * d * cos(θ)

Where:

  • W is the work done (usually measured in Joules).
  • F is the magnitude of the force (usually measured in Newtons).
  • d is the distance over which the force is applied (usually measured in meters).
  • θ (theta) is the angle between the force vector and the direction of displacement.

Now, why is that cos(θ) term there? It accounts for the fact that only the component of the force acting in the direction of motion actually contributes to the work done. If the force is applied at an angle, we need to find the component of the force that's parallel to the displacement. If the force and displacement are in the same direction (θ = 0°), then cos(0°) = 1, and the formula simplifies to W = F * d. If the force is perpendicular to the displacement (θ = 90°), then cos(90°) = 0, and no work is done.

Think about pushing a box across the floor. If you push directly horizontally, all your force contributes to moving the box. But if you push at an upward angle, some of your force is lifting the box, not moving it forward. Only the horizontal part of your push does work in moving the box across the floor.

Understanding this basic concept of work is super important to solve any work-related problem, so make sure to have this concept in mind.

Setting Up the Problem

Alright, let's get back to our problem. We know the block is initially at rest, and we want to find the work done by force F1 over a distance of 8 meters. To solve this, we need a bit more information. Specifically, we need to know:

  1. The magnitude of force F1 (how strong is the force?).
  2. The angle at which force F1 is applied relative to the direction of motion.

Without these crucial details, we can't calculate the work done. Let's assume some values to illustrate how the calculation would work. Suppose:

  • Force F1 has a magnitude of 10 Newtons (F1 = 10 N).
  • Force F1 is applied horizontally, meaning the angle between F1 and the direction of motion is 0 degrees (θ = 0°).

With these assumptions, we can now proceed with the calculation.

Calculating the Work Done

Using the formula for work, W = F * d * cos(θ), we can plug in the values we have:

W = 10 N * 8 m * cos(0°)

Since cos(0°) = 1, the equation becomes:

W = 10 N * 8 m * 1

W = 80 Joules

Therefore, with our assumed values, the work done by force F1 over a distance of 8 meters is 80 Joules. Remember, this is based on our assumption that F1 = 10 N and θ = 0°. If these values are different, the work done will also be different.

Dealing with Angles

What if the force isn't applied horizontally? Let's consider a scenario where force F1 is applied at an angle of 30 degrees to the horizontal (θ = 30°), and the magnitude of F1 is still 10 N. Now, the calculation changes slightly:

W = 10 N * 8 m * cos(30°)

We know that cos(30°) ≈ 0.866, so:

W = 10 N * 8 m * 0.866

W ≈ 69.28 Joules

In this case, the work done is approximately 69.28 Joules. Notice that the work done is less than when the force was applied horizontally. This is because only a component of the force is contributing to the motion in the horizontal direction.

To find the component of the force in the direction of motion, we use F_horizontal = F * cos(θ). In our example, F_horizontal = 10 N * cos(30°) ≈ 8.66 N. This is the effective force that's actually moving the block horizontally.

What if the Block Isn't Initially at Rest?

The problem states that the block is initially at rest. This information is actually important! While the initial velocity doesn't directly appear in the work formula W = F * d * cos(θ), it does influence the overall situation. Here's why:

  • Constant Force and Acceleration: If the force F1 is constant, it will cause the block to accelerate according to Newton's second law (F = ma). If the block starts at rest, its velocity will increase over the 8-meter distance. If the block already had an initial velocity, its final velocity after 8 meters would be different.
  • Work-Energy Theorem: The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy (W_net = ΔKE). Kinetic energy is given by KE = (1/2) * m * v^2, where m is the mass and v is the velocity. So, the work done by F1 (minus any work done by friction, if present) will change the block's kinetic energy. If we knew the block's mass, we could calculate its final velocity after the 8-meter displacement.

In essence, stating that the block starts at rest gives us a starting point for understanding how the force F1 affects its motion. If the block was already moving, the analysis would become more complex, as we'd need to consider the initial kinetic energy.

Considering Friction

In a real-world scenario, friction is almost always present. Friction is a force that opposes motion, and it does negative work. If there's friction between the block and the surface, we need to take that into account when calculating the net work done on the block.

The force of friction is usually given by F_friction = μ * N, where μ is the coefficient of friction (a number between 0 and 1 that represents how rough the surfaces are) and N is the normal force (the force exerted by the surface perpendicular to the block). On a horizontal surface, the normal force is usually equal to the weight of the block (N = mg, where m is the mass and g is the acceleration due to gravity, approximately 9.8 m/s²).

The work done by friction is W_friction = -F_friction * d, where d is the distance. The negative sign indicates that friction does negative work, meaning it takes energy away from the system.

To find the net work done on the block, we subtract the work done by friction from the work done by F1:

W_net = W_F1 - W_friction

The net work is what determines the change in the block's kinetic energy, as stated by the work-energy theorem.

Key Takeaways

  • Work is done when a force causes displacement. The formula is W = F * d * cos(θ). Don't forget to account for the angle between the force and displacement!
  • The initial velocity matters. While it doesn't directly appear in the work formula, it affects the change in kinetic energy and the final velocity of the object.
  • Friction opposes motion and does negative work. Remember to include friction in your calculations to find the net work done.
  • The work-energy theorem connects work and kinetic energy. The net work done on an object equals its change in kinetic energy.

By understanding these concepts and applying them carefully, you can solve a wide range of work-related problems in physics. Keep practicing, and you'll become a pro in no time!

Conclusion

So, to wrap things up, calculating the work done by a force like F1 involves understanding the basics of work, considering the angle of the force, and accounting for factors like friction. Remember to identify all the forces acting on the object and their respective directions to accurately determine the net work done. Keep practicing, and soon you'll be a work-calculation wizard! Good luck, and keep exploring the fascinating world of physics!