JEE Main 2025 Physics question
A force acts on a particle in a plane x + y = 10. The work done by this force during a displacement from (0, 0) to (4m, 2m) is Joule (round off to the nearest integer)
Answer: 152
Solution
Integrating from 0 to 4 and from 0 to 2, sum gives 152 J.