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

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    2021-07-22T16:02:44+00:00

    Answer:

    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  1.40 \hat i+7.20 \hat j.

    The displacement of the coin is

    \vec d=1.40 \hat i+7.20 \hat j

    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

    \vec F=(4.50 cos 128^\circ)\hat i+(4.50 sin 128^\circ)\hat j

       =-2.77 \hat i+3.56 \hat j

    We know that,

    work done is a scalar product of force and displacement.

    W=\vec F.\vec d

        =(-2.77 \hat i+3.56 \hat j).(1.40 \hat i+7.20 \hat j)

        =(-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.

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