Work done, W =F⋅s Here s=rf−ri=(2i^−j^) W=(5i^+3j^+2k^)(2i^−j^)=10−3=7 J
A force F=(5 i+3 j+2 k) N is applied over a particle which displaces it from its origin to the point =(2 i- j) m. The work done on the particle in…
JEE Main 2004 — Physics Mechanics
A force F=(5i^+3j^+2k^)N is applied over a particle which displaces it from its origin to the point r=(2i^−j^)m. The work done on the particle in joules is
Held on 30 Apr 2004 · Verified 6 Jul 2026.
Options
- A
−7
- B
+7
- C
+10
- D
+13
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