By the definition of the work done, we have:
W=FsLet's first find displacements of the body at time t1=2 s and t2=5 s:
d1=(2⋅2i^−4⋅2j^)=(4i^−8j^) m,d2=(2⋅5i^−4⋅5j^)=(10i^−20j^) m.The displacement of the body between t1=2 s and t2=5 s can be calculated as follows:
d=d2−d1=(10i^−20j^)−(4i^−8j^)=(6i^−12j^) m. Finally, we can find the work done between t1=2 s and t2=5 s:
W=(30i^+40j^) N⋅(6i^−12j^) m,W=(30,40)⋅(6,−12)=30⋅6+40⋅(−12)=−300 J.
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