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pantera1 [17]
2 years ago
6

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.
Physics
1 answer:
Ilya [14]2 years ago
3 0

Answer:

Work done, W = 3.78 Joules

Explanation:

It is given that,

Displacement in x-y coordinate, d=(4i+5.8j)\ m

Magnitude of force, F = 1.3 N

The force is directed at a counterclockwise angle of 121° from the positive direction of the x axis. Let W is the work done by the force on the coin during the displacement.

Firstly, finding the vector form of force as :

F=1.3\ cos(121)i+1.3\ sin(121)j

F=-0.669i+1.114j

The work done is given by :

W=F.d

W=(-0.669i+1.114j).(4i+5.8j)

W=-0.669\times 4+1.114\times 5.8

W = 3.78 Joules

So, the work done by the force is 3.78 Joules. Hence, this is the required solution.

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