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A constant force equal to F=3i+2j N is applied to a block which weighs 3.0 kg....

A constant force equal to F=3i+2j N is applied to a block which weighs 3.0 kg. The block is moved from a point <1.2,1.4,0> and eventually arrives at a position <1.9,1.1,0>.

a) How much work is done by the force F given above?
b) What is the change in kinetic energy of the block?
c) If the block starts from 3.0 m/s what is the magnitude of its final velocity?

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Answer #1

Solutionn -=1.2i + 1.4j 1.9i +1.1j (191+11)-1.2i+14j) -0.7i-0.3j a) Work done-FAF (Bi+2) (0.7-03) -2.1-0.6 Therefore the workWork doneinital EWork done +iital 1.5+_x3x3 -1.5+13.5 -15J Magnitude of final velocity is mit15 15 x2 15x2 - 2.83m/s So, the

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