Question

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.

Answer 1

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:

$\Delta U=-5 J$ is the loss in potential energy of the system

$\Delta K=+6 J$ 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):

$E=K+U$

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

$\Delta E= \Delta K + \Delta U$

In this problem:

$\Delta K=+6 J$

$\Delta U=-5 J$

So, the change in mechanical energy is:

$\Delta E=+6+(-5)=+1 J$

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

$W=\Delta E=+1 J$

Answer 2

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

The given parameters;

• change in the total kinetic energy, ΔK.E = 6 J
• change in the initial potential energy, ΔU = - 5J

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

$\Delta W_i + \Delta W_e = \Delta K.E _{tot}$

Where;

$\Delta W_i$ is the work done by the internal forces

$\Delta W_e$ is the work done by the external forces

$\Delta W_e = \Delta U_i \ + \ \Delta _{K.E}\\\\\Delta W_e = -5 + 6\\\\\Delta W_e = + 1 \ J$

The work done by the external forces = 1 J

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

$\Delta W_i + 1 = 6\\\\\Delta W_i= 6-1 \\\\\Delta W_i= 5 \ J$

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

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