A coin slides over a frictionless plane and across an xy coordinate system from

Question

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (4.0 m, 5.8 m) while a constant force acts on it. The force has magnitude 1.3 N and is directed at a counterclockwise angle of 121° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

I understand that W=Fd and F=MA

W=1.3*d, but Im not quite sure how exactly to find displacement.

I would really appreciate it if you could explain how to go about solving this. Thank you!

Explanation / Answer

The coin travels from origin to (4, 5.8) .The displacment is found by using the Pythagorean Theorem and the two sides, 4 and 5.8 to find a hypotenuse length of 7.05 The angle a that this line makes with the + x-axis is given by tan a =5.8/4. Taking Inv Tan on a calculator gives a =54.462 deg. The constant force F acts at an angle of 121 deg The angle ? between the force and the displacment vectors is thus 121 - 54.4, so ? = 66.6 deg. The defn of Work is W = F d cos ? = (1.3N) (7.05 m) cos 66.6 = 3.6 J !!!!!!!! final answer

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