Question #159431

Calculate the work done in displacing a body with a force,f=(30î + 40j)N through S=(2tî-4tj)m between t1=2 second and t2=5 seconds


1
Expert's answer
2021-01-29T02:25:30-0500

By the definition of the work done, we have:


W=FsW=Fs

Let's first find displacements of the body at time t1=2 st_1=2\ s and t2=5 st_2=5\ s:


d1=(22i^42j^)=(4i^8j^) m,d_1=(2\cdot2\hat{i}-4\cdot2\hat{j})=(4\hat{i}-8\hat{j})\ m,d2=(25i^45j^)=(10i^20j^) m.d_2=(2\cdot5\hat{i}-4\cdot5\hat{j})=(10\hat{i}-20\hat{j})\ m.

The displacement of the body between t1=2 st_1=2\ s and t2=5 st_2=5\ s can be calculated as follows:


d=d2d1=(10i^20j^)(4i^8j^)=(6i^12j^) m.d=d_2-d_1=(10\hat{i}-20\hat{j})-(4\hat{i}-8\hat{j})=(6\hat{i}-12\hat{j})\ m.

Finally, we can find the work done between t1=2 st_1=2\ s and t2=5 st_2=5\ s:


W=(30i^+40j^) N(6i^12j^) m,W=(30\hat{i}+40\hat{j})\ N\cdot(6\hat{i}-12\hat{j})\ m,W=(30,40)(6,12)=306+40(12)=300 J.W=(30, 40)\cdot(6,-12)=30\cdot6+40\cdot(-12)=-300\ J.

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