Question

Object $A$is stationary while objects $BandC$are in motion. Forces from object $Ado10J$of work on object $Band–5J$of work on object $C$. Forces from the environment do $4J$of work on object $Band8J$of work on object $C$. Objects $BandC$do not interact. What are $∆Ktotand∆Uint$if

(a) objects $A,B,andC$are defined as separate systems and

(b) one system is defined to include objects $A,B,andC$and their interactions?

Short answer

(a) Total change in kinetic energy$∆k_{total=}17Jand∆U_{internal}=0$.

(b) The change in total kinetic energy and internal energy is $\Delta K_{\text{total}}=12\mathrm{J}$and

$\Delta U_{\text{intemal}}=5\mathrm{J}$respectively.

Step-by-step solution

Introduction

A system is a collection of related items that act together as pieces of an integrated network or process.

Explanation (a)

Since$\mathrm{A},\mathrm{B}$, and $\mathrm{C}$are defined as distinct systems, add $\Delta K$for each system to get the total. If $\mathrm{A}$is stationary then its $K$does not change. Therefore,

$\Delta K_{\mathrm{A}}=0$

According to the work energy theorem, the force of A on B increases $K_{\mathrm{B}}$by $10\mathrm{J}$and the forces due to environment increase it by $4\mathrm{J}$. Therefore,

$\Delta K_{\mathrm{B}}=10\mathrm{J}+4\mathrm{J}$

$=14\mathrm{J}$

The force of A on C decreases its kinetic energy by $5\mathrm{J}$and the forces due to the environment increase it by $8\mathrm{J}$. Therefore,

$\Delta K_{\mathrm{C}}=8\mathbf{J}-5\mathrm{J}$

$=3\mathrm{J}$

The sum of all these change in energy is calculated as follows,

$\Delta K_{\text{weal}}=\Delta K_{\mathrm{A}}+\Delta K_{\mathrm{B}}+\Delta K_{\mathrm{c}}$

$=0\mathrm{J}+14\mathrm{J}+3\mathrm{J}$

$=17\mathrm{J}$

All of the effort is done to change the kinetic energy of the individual systems because all of the forces are external to the system.

As a result, A, B, and C have the same internal energy.

Therefore,

All of the effort is done to change the kinetic energy of the individual systems because all of the forces are external to the system.

As a result, A, B, and C have the same internal energy.

Therefore,

$\Delta U_{\text{internal}}=0$

Therefore, total change in kinetic energy is $\Delta K_{\text{wotal}}=17\mathrm{J}$and $\Delta U_{\text{intemal}}=0$.

Explanation (b)

When all three items are considered as one system, the interaction energy increases owing to object interaction. $\mathrm{A}$$onBandC$. Therefore, change in total internal energy is calculated as follows,

$\Delta U_{\text{intemal}}=10\mathrm{J}-5\mathrm{J}$

$=5\mathrm{J}$

In this case, the kinetic energy is increased by the work done by external forces (environment).

Therefore,

$\Delta K_{\text{lotal}}=8\mathrm{J}+4\mathrm{J}$

$=12\mathrm{J}$

Hence, the change in total kinetic energy and internal energy is $\Delta K_{\text{total}}=12\mathrm{J}$and

$\Delta U_{\text{intemal}}=5\mathrm{J}$respectively.