Question

# To keep her dog from running away while she talks to a friend, Susan pulls gently...

To keep her dog from running away while she talks to a friend, Susan pulls gently on the dog's leash with a constant force given by F⃗  = (2.3 N )x^ + (1.4 N )y^.

Part A

How much work does she do on the dog if its displacement is d⃗  = (0.35 m )x^,

Part B

How much work does she do on the dog if its displacement is d⃗  = (0.20 m )y^,or

Part C

How much work does she do on the dog if its displacement is d⃗  = (-0.50 m )x^ + (-0.30 m )y^?

Work done is given by the Dot product of force & displacement.

And we know that x^.y^ is zero since they both are perpendicular.

And x^.x^ = 1 & y^.y^ = 1

Now

Part A)

Work done = {(2.3 N )x^ + (1.4 N )y^}.(0.35x^ )

Work = 2.3x^.(0.35x^) + 1.4y^.(0.35 x^)

Work = 0.805 Joules

Part B)

Work = {(2.3 N )x^ + (1.4 N )y^}.(0.2y^)

Work = 2.3x^. (0.2y^) + 1.4y^. (0.2y^)

Work = 0.28 Joules

Part C)

Work = {(2.3 N )x^ + (1.4 N )y^}.{(-0.50 m )x^ + (-0.30 m )y^}

Work = 2.3x^.(-0.5x^) + 2.3x^. (-0.3y^) + 1.4y^.(-0.5x^) + 1.4y^.(-0.3y^)

Work = (-1.15) + (-0.42)

Work = (-1.57) Joules

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