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A force is dependent on position and is given by (4.00 N/m) x +(2.0 N/m 2) xy . An object begins at the origin. It first moves in a straight line to x = 1.00 m, y = 0.00 m. It then moves in a straight line to x = 1.00 m, y = 1.00 m. How much work is done on the object by the force during the motion described?

Respuesta :

Answer:

[tex]W=6\sqrt{2} \ J[/tex]

Explanation:

Given:

expression of force:

[tex]F=4x+2xy[/tex]

initial position of the object, [tex]x_o=0;~~~y_o=0[/tex]

position after the first move, [tex]x'=1\ m;~~~y=1\ m[/tex]

final position after the last move, [tex]x=1\ m;~~~y=1\ m[/tex]

As we know that work is defined when the force is applied and the body moves in the direction of force.

The total displacement of the object:

[tex]s=\sqrt{2}[/tex]

Now the force from initial point to the final point in the direction of displacement :

[tex]F_{net}=F-F_o[/tex]

[tex]F_{net}=(4+2)-(0+0)[/tex]

[tex]F_{net}=6\ N[/tex]

Now work done:

[tex]W=F.s\\[/tex]

[tex]W=6\times \sqrt{2}[/tex]

[tex]W=6\sqrt{2} \ J[/tex]

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