A small particle moves to position 5 i ^ - 2 j ^ + k ^ from its initial position 2 i ^ + 3 j ^ - 4 k ^ under…

A small particle moves to position 5i^-2j^+k^ from its initial position 2i^+3j^-4k^ under the action of force 5i^+2j^+7k^ N. The value of work done will be ______ J.

Solution

Work done by a constant force is given by W=ForceF·Displacementr, where, r=r2-r1

Given here, r1=5i^-2j^+k^ and r2=2i^+3j^-4k^

So, W=F·r2-r1

=(5i^+2j^+7k^)·(5i^-2j^+k^)-(2i^+3j^-4k^)

W=40 J

Asked in: JEE Main 2023 (01 Feb Shift 1)

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