The work done by an applied variable force F = x + x³ from x = 0 m to x = 2 m, where x is displacement , is
Displacement is a vector quantity that measures the shortest straight-line distance between an object’s initial and final positions. It includes both magnitude and direction, distinguishing it from distance. Displacement is essential in motion analysis as it represents the change in position regardless of the path taken by the object.
Chapter 5 of Class 11 Physics explores Work Energy and Power. Work involves energy transfer when a force moves an object. Energy appears in forms like kinetic and potential and follows the law of conservation. Power defines the rate of doing work and these principles are vital in understanding mechanics.
Share
To calculate the work done by the variable force F = x + x³ over the displacement from x = 0 m to x = 2 m, we use the work integral:
Work W = ∫[0 to 2] F(x) dx
Where:
– F(x) = x + x³
– x₁ = 0, x₂ = 2
Step 1: Set up the integral
W = ∫[0 to 2] (x + x³) dx
Step 2: Split and evaluate the integral
W = ∫[0 to 2] x dx + ∫[0 to 2] x³ dx
1. For ∫[0 to 2] x dx:
∫ x dx = (1/2) x² |[0 to 2] = (1/2) (2²) – (1/2) (0²) = 2
2. For ∫[0 to 2] x³ dx:
∫ x³ dx = (1/4) x⁴ |[0 to 2] = (1/4) (2⁴) – (1/4) (0⁴) = (16/4) – 0 = 4
Step 3: Add the results
W = 2 + 4 = 6 J
Final Answer:
The work done by the variable force is 6 J.
For more information:
https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/