Q.
A body moves from a position r1=(2i^−3j^−4k^)m to a position, r2=(3i^−4j^+5k^)m under the influence of a constant force F=(4i^+j^+6k^)N . The work done by the force is
3120
217
NTA AbhyasNTA Abhyas 2020Work, Energy and Power
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Solution:
Given, r1=2i^−3j^−4k^
And r2=3i^−4j^+5k^
Now, r2−r1=i^−j^+9k^
And F=4i^+j^+6k^ ∴ work done = F.r W=(4i^+j^+6k^).(i^−j^+9k^ ) =4−1+54=57J