Question detail
Forces and elasticity scenario: a velocity-time graph shows acceleration, steady speed, and deceleration. Which answer best addresses Elastic potential energy and the objective to interpret force-extension graphs in terms of energy stored where appropriate?
Try the question, check the answer, then read the explanation to understand the curriculum point.
At a glance
MCQ
Type
practice
Style
Topic
Forces and elasticity
Question
- A. In the velocity graph scenario, apply extension to interpret force-extension graphs in terms of energy stored where appropriate while keeping elastic versus plastic deformation separate.
- B. In the velocity graph scenario, mix up elastic versus plastic deformation and ignore extension.
- C. Use a general revision statement without applying Elastic potential energy to the situation.
- D. Choose a different forces topic instead of explaining interpret force-extension graphs in terms of energy stored where appropriate.
Answer
The correct answer is: In the velocity graph scenario, apply extension to interpret force-extension graphs in terms of energy stored where appropriate while keeping elastic versus plastic deformation separate.
Explanation
The correct option is In the velocity graph scenario, apply extension to interpret force-extension graphs in terms of energy stored where appropriate while keeping elastic versus plastic deformation separate.. It is correct because the scenario says a velocity-time graph shows acceleration, steady speed, and deceleration, which must be interpreted through Elastic potential energy. This directly supports the learning objective to interpret force-extension graphs in terms of energy stored where appropriate. Use values 7, 11, and 17 only if the question asks for a calculation. The answer earns credit by naming the relevant force or motion quantity, using units when needed, and avoiding the boundary error elastic versus plastic deformation.
Common mistake
Misinterpreting the slope as energy
Students often think the slope of a force‑extension graph directly gives the elastic potential energy stored, rather than the spring constant.
Explain that the slope is the spring constant (k) and that elastic potential energy is calculated using 0.5 k e², where e is the extension in metres.
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