JEE Main202524 Jan 2025Morning ShiftPhysicsWork, Power and EnergyActual
A force F = + x ^2 acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant =1 ~N then will be
Options
- A15 ~N / m ^2
- B12 ~N / m ^2
- C8 ~N / m ^2
- D10 ~N / m ^2
Correct answer
B. 12 ~N / m ^2
Step-by-step solution
F= + x^2 Work done d w= F d x aligned & W= F d x= ( + x^2 ) d x & W= | x+ x^3 3 |₀^1= + 3 =5 aligned Given =1 So, 3 =4 =12 ~N / m ^2