Calculating Work Done: Physics Problem Solved!
Hey guys! Let's dive into a classic physics problem. We've got a body with a mass of 2 kg, and a constant force is applied to it. The velocity of the body changes over time, as shown in the graph. The big question: what's the work done in the first 10 seconds? Don't worry, we'll break it down step-by-step, so you can totally nail it! We will explore the calculation of work in physics problems, breaking down each step to make it super clear and easy to understand. We'll be using fundamental concepts like Newton's Second Law of Motion, the relationship between work, force, and displacement, and how to interpret velocity-time graphs. This problem is a great example of how different physics principles come together. So, grab your calculators and let's get started. We'll show you how to find the work done. This is useful for anyone studying physics, whether you're in high school or college, or just curious about how things move. By the end, you'll be able to calculate work done. Let's start with a breakdown of the problem.
Firstly, we have a body with a mass (m) of 2 kg. We know that a constant force is acting on it. The tricky part? We are given a velocity-time graph – this is key! The graph tells us how the velocity changes over time. Understanding this is crucial. In physics, the work (W) done on an object is related to the force applied and the distance the object moves. The formula is: W = F * d (Work = Force * Distance). However, in this problem, we aren't directly given the force or the distance. We need to figure them out using the information we have, which is the mass and the velocity-time graph. The goal is to calculate the work done in the first 10 seconds, and we must find the force and the displacement of the object within this time interval. But hold your horses! Before we jump into calculations, we need to gather all the relevant information and understand the relationships between the different physical quantities involved. We'll use the graph to determine the object's acceleration and then use Newton's Second Law to find the force. Once we have the force, we can find the distance traveled using the velocity-time graph. So let's crack on and figure this out together.
Decoding the Velocity-Time Graph
Alright, let's get our detective hats on and analyze the velocity-time graph. This is where the fun begins. The graph shows us how the velocity of the object changes over time. We need to extract useful information from this graph. Look closely; the graph gives us the velocity of the object at different points in time. The slope of the velocity-time graph represents the object's acceleration (a). Acceleration is the rate at which velocity changes. If the graph is a straight line, it means the acceleration is constant. In this case, we have a straight line, which simplifies things. In a velocity-time graph, the slope is constant, which signifies constant acceleration. This allows us to use simple kinematic equations to solve the problem. The area under the velocity-time graph gives us the displacement (d), which is the distance traveled by the object during a specific time interval. The displacement is how far the object has moved. Let's find the acceleration and the displacement using the graph. The acceleration can be found by calculating the slope of the line. The slope (acceleration) can be calculated by dividing the change in velocity by the change in time. The formula is: a = (v2 - v1) / (t2 - t1). Here, v2 is the final velocity, v1 is the initial velocity, t2 is the final time, and t1 is the initial time. Looking at the graph, the initial velocity (v1) at t1 = 0 s is 0 m/s. At t2 = 10 s, the final velocity (v2) is 20 m/s. So, let's plug these values into our formula. The acceleration will be: a = (20 m/s - 0 m/s) / (10 s - 0 s) = 2 m/s². The object has a constant acceleration of 2 m/s². That's great information. Now we can proceed.
To find the displacement, we need to calculate the area under the graph for the first 10 seconds. In this case, the area under the graph is a triangle, as the velocity increases linearly from 0 to 20 m/s over 10 seconds. The area of a triangle is given by (1/2) * base * height. Here, the base is the time interval (10 s), and the height is the final velocity (20 m/s). So the displacement will be: d = (1/2) * 10 s * 20 m/s = 100 m. Therefore, the object traveled a distance of 100 meters in the first 10 seconds. Now that we have the acceleration and displacement, we can calculate the force and the work done. We are well on our way to solving this problem. Keep in mind that understanding how to extract information from graphs is super important in physics! Keep in mind that we're going to use all this info in the next step to figure out the work done.
Calculating the Force and Work Done
Awesome, now that we've found the acceleration and displacement, we can calculate the force and finally the work done. Let's start with the force. Remember Newton's Second Law of Motion: F = m * a (Force = mass * acceleration). We know the mass (m) of the object is 2 kg, and we've calculated the acceleration (a) to be 2 m/s². Now we just plug in the values and solve. The force will be: F = 2 kg * 2 m/s² = 4 N. The force acting on the object is 4 Newtons. Nice! Now that we have the force, we can easily calculate the work done using the formula W = F * d. We know the force (F) is 4 N, and the displacement (d) is 100 m. So, let's calculate the work done: W = 4 N * 100 m = 400 Joules. The work done on the object in the first 10 seconds is 400 Joules. And there you have it! We've solved the problem. We started with the mass and the graph, calculated the acceleration and displacement, found the force, and finally calculated the work done. Easy peasy, right?
To recap: We’ve successfully found the work done, and this process illustrates the practical application of physics principles. We will make it super easy for you to understand, no matter your background. Remember that the work done represents the energy transferred to the object due to the force causing it to move over a distance. This is also a good example of how to break down complex problems into smaller, manageable steps. Remember the key formulas: F = m * a and W = F * d. And always remember to pay close attention to your graphs – they’re full of valuable information! Now that you’ve seen how it's done, try solving similar problems. That's the best way to get better! Keep practicing these types of problems, and you'll become a physics pro in no time! Keep up the amazing work!
Summary of Formulas Used
Here's a quick recap of the formulas we used:
- Acceleration: a = (v2 - v1) / (t2 - t1)
- Newton's Second Law: F = m * a
- Work Done: W = F * d
- Area of a Triangle: (1/2) * base * height
Conclusion
We successfully calculated the work done on the object using the provided information and a bit of physics know-how. This problem highlights the importance of understanding how to interpret graphs and apply fundamental physics principles. Keep practicing, and you'll get better! If you have any questions, feel free to ask. And remember, physics is all about understanding how things work, so embrace the challenge and have fun with it!