A particle experiences a variable force F⃗ = ( 4xî + 3y²ĵ ) in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by :

Solution

Physics sir · 4 steps

The force changes with position, so integrate each component along its own axis and add the two works.

1. Work-energy theorem. The plane is horizontal, so gravity and the normal force do no work. F⃗ is the only force doing work, so Δ K = W = ∫ F⃗ · dr⃗.

   W = ∫₁² 4x dx + ∫₂³ 3y² dy

2. x-part. ∫₁² 4x dx = [2x²]₁² = 8 - 2 = 6 J.
3. y-part. ∫₂³ 3y² dy = [y³]₂³ = 27 - 8 = 19 J.
4. So. Δ K = 6 + 19 = 25 J, which is option (c).

The idea to keep. When Fₓ depends only on x and F_y only on y, the work splits into separate integrals. The path does not matter, only the start and end points.

Answer from the NTA final answer key. Solution by PrepWiser.