NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A force F is related to the position of a particle by the relation F = ( 10 x 2 ) N . The work done by the force when the particle moves from x = 2 m to x = 4 m is
Options
- A5 6 3 J
- B560   J
- C5 6 0 3   J
- D3 5 6 0 J
Correct answer
C. 5 6 0 3   J
Step-by-step solution
W = ∫ 2 4 Fdx = ∫ 2 4 10 x 2 d ⁡ x = 10 x 3 3 2 4 = 1 0 3 4 3 - 2 3 = 1 0 3 × 5 6 = 5 6 0 3    J