Conservation of Energy — Question 3

Section 8-7 · Physics pages 1136–1137

Question 3 An internal battery can increase mechanical energy A system includes a battery-powered lifting mechanism, a load and Earth. No work or heat crosses its boundary. During one operation, battery chemical energy decreases 100 J, gravitational potential increases 60 J, and total macroscopic kinetic energy increases 10 J. No other store changes except thermal energy. battery motion / height thermal All arrows represent transfers between stores inside one closed boundary. (a) Find the thermal-energy increase. (b) Find the change in mechanical energy. (c) Explain why zero external work does not require mechanical energy to remain constant.
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Question 3 - Solution An internal battery can increase mechanical energy battery motion / height thermal (a) Sum changes in all stores. Sum every storage change inside the isolated boundary: 0 = −100 + 60 + 10 + ∆Eth ⇒∆Eth = 30 J . (b) Use the definition of mechanical energy. Mechanical energy comprises kinetic and potential stores: ∆Emech = ∆K + ∆Ug = 10 + 60 = 70 J . (c) Identify the internal source. The internal source supplies both outputs: −∆Echem = ∆Emech + ∆Eth = 70 + 30 = 100 J. Zero boundary transfer conserves the total, not each energy category. Omitting the battery would omit the source of the mechanical increase.

Question 3

An internal battery can increase mechanical energy

A system includes a battery-powered lifting mechanism, a load and Earth. No work or heat crosses its boundary. During one operation, battery chemical energy decreases 100 J, gravitational potential increases 60 J, and total macroscopic kinetic energy increases 10 J. No other store changes except thermal energy.

Diagram shown in the original SVG worksheet.

  1. Find the thermal-energy increase.

  2. Find the change in mechanical energy.

  3. Explain why zero external work does not require mechanical energy to remain constant.

Question 3 - Solution

An internal battery can increase mechanical energy

Diagram shown in the original SVG worksheet.

(a) Sum changes in all stores.

Sum every storage change inside the isolated boundary: 0=−100+60+10+ΔEth⇒ΔEth=30J.0=-100+60+10+\Delta E_{\mathrm{th}}\Rightarrow\boxed{\Delta E_{\mathrm{th}}=30\ \mathrm J}.

(b) Use the definition of mechanical energy.

Mechanical energy comprises kinetic and potential stores: ΔEmech=ΔK+ΔUg=10+60=70J.\boxed{\Delta E_{\mathrm{mech}}=\Delta K+\Delta U_g=10+60=70\ \mathrm J}.

(c) Identify the internal source.

The internal source supplies both outputs: −ΔEchem=ΔEmech+ΔEth=70+30=100J.-\Delta E_{\mathrm{chem}}=\Delta E_{\mathrm{mech}}+\Delta E_{\mathrm{th}}=70+30=100\ \mathrm J. Zero boundary transfer conserves the total, not each energy category. Omitting the battery would omit the source of the mechanical increase.