Respuesta :
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 [tex]1.40 \hat i+7.20 \hat j[/tex].
The displacement of the coin is
[tex]\vec d=1.40 \hat i+7.20 \hat j[/tex]
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
[tex]\vec F=(4.50 cos 128^\circ)\hat i+(4.50 sin 128^\circ)\hat j[/tex]
[tex]=-2.77 \hat i+3.56 \hat j[/tex]
We know that,
work done is a scalar product of force and displacement.
[tex]W=\vec F.\vec d[/tex]
[tex]=(-2.77 \hat i+3.56 \hat j).(1.40 \hat i+7.20 \hat j)[/tex]
=(-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.
