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?

Answer 1

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.