Homeworkify AI Homework Helper , Scan Solve Learn : Physics Demo
TLDRIn this Physics tutorial, we tackle a problem involving work done by a force. Given a force of 15N at a 45-degree angle to the horizontal and a displacement of 49 meters, we calculate the work by finding the horizontal component of the force and applying it to the formula for work. The result is 520.89 Joules, which converts to 0.1244 kilo calories, demonstrating the work done in moving an object horizontally over the given distance.
Takeaways
- 📐 The problem involves calculating work done by a force with a given force magnitude, angle, and displacement.
- 🧮 The formula for work done by a force is given by `work = force x displacement x cos(angle)`.
- 🔍 The angle between the force and the displacement is 45 degrees.
- 📏 The horizontal component of the force is calculated using trigonometry: `horizontal component = force x cos(angle)`.
- 📉 The horizontal component of the force is found to be 10.61 N.
- ⚙️ The work done by the force is calculated with the horizontal component and the displacement: `work = 10.61 N x 49 m x cos(0 degrees)`.
- 📈 The work done is 520.89 Joules, since the angle between the horizontal direction and displacement is 0 degrees.
- ↔️ To convert Joules to kilo calories (KCAL), use the conversion `1 KCAL = 4184 J`.
- 🔗 The work done in KCAL is found by dividing the work in Joules by 4184, resulting in 0.1244 KCAL.
- 🔄 The work done by the force in moving an object horizontally over a distance of 49 meters is equivalent to 0.1244 kilo calories.
- ❓ The speaker encourages asking further questions for clarification or additional physics problems.
Q & A
What is the given force in the physics problem presented?
-The given force in the problem is 15 Newtons (15N).
What is the angle between the force and the horizontal direction?
-The angle between the force and the horizontal direction is 45 degrees.
What is the distance over which the force is acting?
-The distance over which the force is acting is 49 meters.
What is the formula for work done by a force?
-The formula for work done by a force is work = force x displacement x cos(angle), where angle is the angle between the force and the displacement.
How do you find the horizontal component of the force?
-You find the horizontal component of the force by multiplying the force by the cosine of the angle, so horizontal component of force = force x cos(angle).
What is the calculated horizontal component of the force in this problem?
-The calculated horizontal component of the force is 10.61 Newtons (10.61N).
How is the work done by the force calculated when the force is acting horizontally?
-When the force is acting horizontally, the work done is calculated as work = horizontal component of force x displacement x cos(0 degrees).
What is the work done by the force in Joules?
-The work done by the force, when moving the object horizontally over a distance of 49 meters, is 520.89 Joules.
How can you convert Joules to kilo calories (KCAL)?
-You can convert Joules to kilo calories by dividing the number of Joules by 4184, since 1 kilo calorie equals 4184 Joules.
What is the equivalent of the work done in kilo calories (KCAL)?
-The work done by the force, equivalent in kilo calories, is 0.1244 KCAL.
Can you provide any additional physics concepts or principles related to this problem?
-This problem involves concepts of force, work, and trigonometry. It demonstrates how the direction of force affects the work done and introduces the idea of force components.
Outlines
📐 Physics Problem: Work Done by a Force
The video script begins by introducing a foreign physics problem that involves calculating the work done by a force. The given information includes a force of 15 Newtons (N) at an angle of 45 degrees to the horizontal and a displacement of 49 meters. The script explains the formula for work done by a force, which is 'work = force x displacement x cos(angle)'. Since the force is not in the direction of displacement, the script calculates the horizontal component of the force using trigonometry, resulting in 10.61 N. It then uses this value to find the work done, which is 520.89 Joules (J), by applying the work formula with the horizontal component and a displacement angle of 0 degrees. Finally, the script provides a conversion from Joules to kilo calories (KCAL), showing that the work done is equivalent to 0.1244 KCAL.
Mindmap
Keywords
💡Force
💡Angle
💡Displacement
💡Work
💡Component of Force
💡Trigonometry
💡Horizontal Component
💡Cosine
💡Joules
💡Kilo calorie
Highlights
The foreign physics problem involves calculating the work done by a force.
Given information includes a force of 15N at a 45-degree angle to the horizontal and a displacement of 49 meters.
The formula for work done is force x displacement x cos(angle).
The force is not in the direction of displacement, so the horizontal component of force is needed.
The horizontal component of force is calculated using trigonometry: 15N * cos(45 degrees).
The horizontal component of force equals 10.61N.
The work done by the force is calculated using the horizontal component and the displacement.
The angle between the horizontal direction and displacement is 0 degrees.
The calculated work done is 520.89 Joules.
The work done is equivalent to 0.1244 kilo calories using the conversion 1 kcal = 4184 Joules.
The problem demonstrates the application of physics principles in calculating work done by a force.
The solution involves understanding the relationship between force, displacement, and the angle of application.
Trigonometry plays a crucial role in finding the horizontal component of the force.
The work done by the force is independent of the angle between the force and the displacement direction.
The final answer provides insight into the energy required to move an object horizontally.
The conversion from Joules to kilo calories is a practical application for understanding energy in different units.
The problem-solving process is step-by-step, making it accessible for learning and understanding.
The demonstration includes a real-world application of physics concepts, enhancing the relevance of the subject.
The transcript offers a clear explanation, making complex physics concepts more digestible.
The inclusion of a musical element adds an engaging touch to the educational content.