Question

How much work is done by a force F = (2x N) i^ + (3 N) j^, with x in meters, that moves a particle from a position ri = (2 m) i^ + (3 m) j^ to a position rf = -(4 m) i^ - (3 m) j^?

Answer 1

The work done by the force F is -12x Nm.

Step-by-step explanation:

To find the work done by the force F, we need to calculate the dot product between F and the displacement vector Δr. The dot product is given by F · Δr. Using the given positions, we can calculate Δr = rf - ri = (-(4 m) - (2 m)i^ + (3 m) - (3 m)j^ = (-6 m)i^ + 0 j^.

Now we can calculate the dot product: F · Δr = (2x N)(-6 m) + (3 N)(0) = -12x Nm.

Therefore, the work done by the force F is -12x Nm.