A body of mass 2 kg moving with velocity of v⃗ᵢₙ = 3 î + 4 ĵ ms⁻¹ enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of 5/3 seconds, then velocity of the body when it emerges from force field is.

Solution

Physics sir · 3 steps

The force acts only along z, so only the z-velocity changes; x and y just carry on unchanged.

1. Find acceleration. By Newton's second law, a = F/m = 6/2 = 3 m s⁻², directed along +k̂.
2. Components separately. No force acts along x or y, so vₓ = 3 and v_y = 4 stay the same. Only v_z changes, starting from 0.

   v_z = 0 + a t = 3 × 5/3 = 5 m s⁻¹

3. So. The body leaves the field with v⃗ = 3î + 4ĵ + 5k̂ m s⁻¹, which is option (d).

The idea to keep. Newton's second law works component by component. A force along one axis changes velocity only along that axis; the other components stay exactly as they were.

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