## A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.40 m, 7.20 m) w

Question

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.40 m, 7.20 m) while a constant force acts on it. The force has magnitude 4.50 N and is directed at a counterclockwise angle of 128.° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

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3 days 2021-07-22T16:01:32+00:00 1 Answers 0 views 0

The work done required on the coin during the displacement is 21.75 w.

Explanation:

Given that,

A coin slides over a friction-less plane i.e friction force = 0.

The co-ordinate of the given point is (1.40 m, 7.20 m).

The position vector of the given point is represented by .

The displacement of the coin is The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.

Then x component of the force = 4.50 cos128°

The y component of the force = 4.50 sin128°

Then the position vector of the force is  We know that,

work done is a scalar product of force and displacement.  =(-2.77×1.40+ 3.56×7.20) w

=21.75 w

The work done required on the coin during the displacement is 21.75 w.