A spring 40 mm long is stretched by the application of a force. If 10 N force is required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is
Force is a physical quantity that causes an object to change its state of motion or rest. It is a vector quantity characterized by both magnitude and direction. Forces can be categorized as contact forces and non-contact forces such as gravitational or electromagnetic. The effects of force include changes in motion shape and acceleration.
Class 11 Physics Chapter 5 focuses on work, energy and power explaining their definitions and interrelations. It includes concepts like kinetic energy potential energy work-energy theorem and conservation of energy. Practical applications and problem-solving enhance understanding of these fundamental principles which are crucial for analyzing mechanical systems and various physical phenomena effectively.
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To solve the problem of calculating work done in stretching a spring, we must refer to the property of the spring as described in Hooke’s Law. In this law, it is defined that the amount of force applied to stretch the spring is proportional to the stretching from its original length. The force increases linearly with the degree of stretching the spring. The work done in stretching the spring is equivalent to the energy stored in it, often referred to as elastic potential energy.
The work done is visualized to be the area under a graph of force extension. Since this relationship between the force and extension is linear, the graph plots as a triangle. The extension of the spring is represented by the base of this triangle, while the height would represent the maximum force required. Therefore, work done is directly proportional to the square of extension.
In this given case, the spring would need a force of 10 N for every millimeter of extension, and it is stretched 40 mm. When we use the formula for work done in a spring, substituting the given values enables us to calculate how much energy is in the spring due to this extension. When using the result of this computation, we obtain a total work done of 8 J, meaning it amounts to the amount of energy required to extend the spring by 40 mm.
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https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/