Question

Physics 12 Question




1) Why is work positive for separating two attractive objects?

2) How can work become negative between two objects? What should happen?

Answer 1

if we consider the work done by the extwernal force, then such work is positive, since both external forces and displacement have the same orientation

Answer 2

Lets think about two charged objects in space

Answer 3

How would work be positive if we separate these two attractive objects?

Answer 4

please look at this drawing:

Answer 5

the initial potential energy is:
\[P{E_i} = \frac{{{Q^2}}}{{d_i^2}}\]

Answer 6

I believe work is positive since force and displacement are in the same direction of each other when we are separating the two liked charges.

Answer 7

final potential energy, is:
\[P{E_f} = \frac{{{Q^2}}}{{d_f^2}} < P{E_i}\]

Answer 8

please you have to specify the force which are doing such work

Answer 9

the work done by the force of the field, namely the electrostatic forces, is:
\[{L_E} = P{E_i} - P{E_f} > 0\]

Answer 10

yes

Answer 11

which is positive

Answer 12

so the work done by the external forces is negative, since:
\[\huge {L_{external}} = - {L_E} < 0\]

Answer 13

but I believe there is an easier way of this.

It has to do something with the direction of force and direction of displacement.

I found this on the internet now: "Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work"

Your explanation was helpful indeed

Answer 14

please keep in mind that the external forces are not separating the objects
the two objects are separating each from other due to the internal forces namely the electrostatic forces

Answer 15

You mean by internal forces such as repulsive forces?

Answer 16

yes!

Answer 17

Ok got it. Good teamwork :D

Answer 18

:)

Answer 19

oops.. I have made a typo, the correct formula for potential energy, is:
\[\huge PE = \frac{{{Q^2}}}{d}\;\left( {CGS} \right)\]