A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answer as an integer. Part B
How much work is done by external forces?
Express your answer as an integer.

The work done by the internal forces is 5 J and the work done by the external forces is 1 J.

The given parameters;

The work done by the external forces and the internal forces is calculated as follows;

Where;

is the work done by the internal forces

is the work done by the external forces

The work done by the external forces = 1 J

The work done by the internal forces is calculated as follows;

Thus, the work done by the internal forces is 5 J and the work done by the external forces is 1 J.

Learn more here:link

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

is the loss in potential energy of the system

is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

In this problem:

So, the change in mechanical energy is:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

a) electrical > light, sound, and thermal< < < <

) the mechanical energy of the firefighter moving.< < < < a) electrical > light, sound, and thermal< < < <

c) both a and b.

iron would move it is magnetic there is a mag field around the current in the wire