Question

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?

Answer 1

Answer:

$W=6\sqrt{2} \ J$

Explanation:

Given:

expression of force:

$F=4x+2xy$

initial position of the object, $x_o=0;~~~y_o=0$

position after the first move, $x'=1\ m;~~~y=1\ m$

final position after the last move, $x=1\ m;~~~y=1\ m$

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:

$s=\sqrt{2}$

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

$F_{net}=F-F_o$

$F_{net}=(4+2)-(0+0)$

$F_{net}=6\ N$

Now work done:

$W=F.s$

$W=6\times \sqrt{2}$

$W=6\sqrt{2} \ J$