A force does work on a 50 g particle as the particle moves along the following straight paths in the xy-plane: 25 J from (0 m, 0

Question

3 years ago

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A force does work on a 50 g particle as the particle moves along the following straight paths in the xy-plane: 25 J from (0 m, 0

m) to (5 m, 0 m); 35 J from (0 m, 0 m) to (0 m, 5 m); -5 J from (5 m, 0 m) to (5 m, 5 m); -15 J from (0 m, 5 m) to (5 m, 5 m); and 20 J from (0 m, 0 m) to (5 m, 5 m).
Is this a conservative force?

If the zero of potential energy is at the origin, what is the potential energy at (5m, 5m)?

Physics

1 answer:

vfiekz [6] · Answer 1 · 2022-03-27T05:46:10+03:00

1) Yes, it is a conservative force

2) The potential energy at (5m, 5m) is 20 J

Explanation:

1)

A force is defined to be conservative if the work done by the force when moving an object does not depend on the path taken, but only on the initial and final position of the object.

Let's verify if this condition is met for the force in this problem:

- For going from (0 m, 0 m) to (5 m, 5 m), the work done is 20 J (A)

Then we can go from (0 m, 0 m) to (5 m, 5 m) through a different path:

- from (0 m, 0 m) to (5 m, 0 m) (work done: 25 J)

- from (5 m, 0 m) to (5 m, 5 m) (work done: -5 J)

Total work done in the second path: 25 J + (-5 J) = 20 J --> same as (A)

We can also go from (0 m, 0 m) to (5 m, 5 m) through a different path:

- from (0 m, 0 m) to (0 m, 5 m) (work done: 35 J)

- from (0 m, 5 m) to (5 m, 5 m) (work done: -15 J)

Total work done in the third path: 35 J + (-15 J) = 20 J --> same as (A)

So, the force is conservative, since the work done does not depend on the path taken.

2)

For a conservative force, the change in potential energy of the object moved by the force is equal to the work done by the force in moving the object from A to B.

Here have:

Point A: (0m, 0m)

Point B: (5m, 5m)

So, the change in potential energy is equal to the work done from A to B:

$W=\Delta U = U_B-U_A$