Question

A force is dependent on position and is given by (4.00 N/m)x+(2.0 N/m2)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

`W=2√(26) \,J`

Step-by-step explanation:

Given that:

Relation of force with position,
`F=4x+2xy` N

Initial position of the object,
`(x_0,y_0)=(0m,0m)`

second position of the object,
`(x_2,y_2)=(1m,0m)`

final position of the object,
`(x_f,y_f)=(1m,1m)`

Now, from the schematic we get the displacement as:

`s=√(2)\,m`

Now we calculate :

The force in X direction for initial displacement

`F_x=4* 1+2* 1* 0`

`F_x=4\,N`

The force in Y direction for initial displacement

`F_y=4* 1+2* 1* 1`

`F_y=6\,N`

Now the resultant force in the direction of displacement:

`F_R=√(F_x\,^2+F_y\,^2)`

`F_R=√(4^2+6^2)`

`F_R=2√(13)\,N`

Therefore work:

`W=F_R* s`

`W=2√(13)* √(2)`

`W=2√(26) \,J`