A student applied a force $$\(F\)$$ to a rubber band. The student measured the corresponding extension, $$\(\Delta x\)$$, as $$\(F\)$$ was increased and then decreased. The graph shows how $$\(\Delta x\)$$ varied as $$\(F\)$$ was increased and then decreased. The work done on the rubber band as $$\(F\)$$ increased is greater than the work done by the rubber band as $$\(F\)$$ was decreased. There is a difference in these values because heating occurred. Determine the energy that caused heating as $$\(F\)$$ was increased and then decreased to zero. (3) .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. Energy that caused heating $$\(=\)$$
Exam No：wph11-01-que-20240112 Year：2024 Question No：11(b)
Answer:
Indication that \(E_{\mathrm{el}}=\) area under force-extension graph.
Uses area between graph lines to determine energy.
Energy \(=0.048 \mathrm{~J}(\) allow a range from 0.0425 J to 0.053 J\()\)
Example of calculation
\(1 \mathrm{~cm}^{2}\) on graph \(\equiv 0.5 \mathrm{~N} \times 0.005 \mathrm{~m}=0.0025 \mathrm{~J}\)
Area between graphs \(\approx 19 \mathrm{~cm}^{2}\)
Energy that caused heating \(=19 \mathrm{~cm}^{2} \times 0.0025 \mathrm{~J} \mathrm{~cm}^{-2}=0.0475 \mathrm{~J}\)
Uses area between graph lines to determine energy.
Energy \(=0.048 \mathrm{~J}(\) allow a range from 0.0425 J to 0.053 J\()\)
Example of calculation
\(1 \mathrm{~cm}^{2}\) on graph \(\equiv 0.5 \mathrm{~N} \times 0.005 \mathrm{~m}=0.0025 \mathrm{~J}\)
Area between graphs \(\approx 19 \mathrm{~cm}^{2}\)
Energy that caused heating \(=19 \mathrm{~cm}^{2} \times 0.0025 \mathrm{~J} \mathrm{~cm}^{-2}=0.0475 \mathrm{~J}\)
Knowledge points:
CH2 - Materials
Solution:
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