JEE Main 2023 · 1 February · morning · Numerical answer · +4 / −0
A small particle moves to position 5 î-2 ĵ+k̂ from its initial position 2 î+3 ĵ-4 k̂ under the action of force 5 î+2 ĵ+7 k̂ N. The value of work done will be __________ J.
Answer and solution
Answer 40
Solution
Physics sir · 3 steps
Work is force dotted with displacement, so find the displacement vector first, final minus initial.
- Find displacement. Subtract the initial position from the final one: d⃗ = (5-2)î + (-2-3)ĵ + (1-(-4))k̂ = 3î - 5ĵ + 5k̂ m.
W = F⃗·d⃗ = Fₓ dₓ + F_y d_y + F_z d_z
- Take the dot product. Multiply matching components and add: W = (5)(3) + (2)(-5) + (7)(5) = 15 - 10 + 35.
- So. The work done is W = 40 J, so the answer is 40.
The common slip: -40. Taking initial minus final position flips every component of the displacement and gives -40 instead of 40. Displacement is always final minus initial.
The idea to keep. For a constant force, W = F⃗·Δr⃗. Always use the displacement (final minus initial), never a position vector on its own, and watch the signs of each component.
Answer from the NTA final answer key. Solution by PrepWiser.