A 1 kg block subjected to two simultaneous forces (2 i + 3 j + 4 k) N and (3 i - j - 2 k) N is moved a distance of 25 m along (3 i - 4 j) direction.…
JEE Main 2026 — Physics Mechanics
2026integermedium
A 1 kg block subjected to two simultaneous forces (2i^+3j^+4k^) N and (3i^−j^−2k^) N is moved a distance of 25 m along (3i^−4j^) direction. The work done in this process is _____ J.
Official previous-year question
Held on 4 Apr 2026 · Verified 6 Jul 2026.
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Solution
The net force acting on the block is the vector sum of the two forces:
F=F1+F2=(2i^+3j^+4k^)+(3i^−j^−2k^)=5i^+2j^+2k^ N
The block is moved a distance of 25 m along the direction of the vector 3i^−4j^. The unit vector along this direction is:
n^=32+(−4)23i^−4j^=53i^−4j^
The displacement vector s is the magnitude multiplied by the unit vector:
s=25n^=25(53i^−4j^)=15i^−20j^ m
The work done W is the dot product of the net force and the displacement vector:
W=F⋅s=(5i^+2j^+2k^)⋅(15i^−20j^)
W=(5×15)+(2×−20)+(2×0)
W=75−40=35 J
Answer: 35
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