A boat of mass 120 kg floats in a stream. You push on the boat with a pole, exerting a force of (175 i hat - 145 j hat) N. The boat moves through a displacement dvec = (23.0 i hat - 8.0 j hat) m along a straight line.

How much work does the force of the pole do on the boat during the displacement?

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To find the work done on the boat, we can use the equation:

Work = Force ∙ Displacement ∙ cos(θ),

where Force is the applied force on the boat, Displacement is the displacement vector, and θ is the angle between the Force and Displacement vectors.

First, let's find the magnitude of the Force vector:

|Force| = √((175^2) + (-145^2)) = √(30625 + 21025) = √(51650) = 227.34 N.

Next, let's find the angle θ between the Force and Displacement vectors. We can use the dot product formula:

θ = cos^(-1)((Force ∙ Displacement) / (|Force| ∙ |Displacement|)).

The dot product of the Force and Displacement vectors is given by:

Force ∙ Displacement = (175 ∙ 23) + (-145 ∙ -8) = 4025 + 1160 = 5185.

The magnitudes of the Force and Displacement vectors are:

|Force| = 227.34 N,
|Displacement| = √((23^2) + (-8^2)) = √(529 + 64) = √(593) = 24.34 m.

Plugging these values into the equation for θ:

θ = cos^(-1)((5185) / (227.34 ∙ 24.34)) = cos^(-1)(5185 / 5539.69) = cos^(-1)(0.935).

Using a calculator, we find:

θ ≈ 20.20 degrees.

Finally, we can plug the values of the Force, Displacement, and θ into the equation for work:

Work = (175 ∙ 23 ∙ cos(20.20 degrees)) + (-145 ∙ -8 ∙ cos(20.20 degrees)) = (4025 ∙ 0.9397) + (1160 ∙ 0.9397) = 3779.47 + 1089.52 = 4868.99 J.

Therefore, the work done on the boat is approximately 4868.99 Joules.