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Physics Chapter 7 : Topic 7
Elastic Potential Energy
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Elastic potential energy is the energy stored in a material when it is:
Reasoning: Elastic potential energy is stored when a material is stretched or compressed and can return to its original shape.
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The elastic potential energy stored in a stretched wire is equal to:
Reasoning: The work done in stretching the wire is stored as elastic potential energy, and work is represented by the area under the graph.
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For a material obeying Hooke’s law, the force-extension graph is:
Reasoning: Within the limit of proportionality, force is directly proportional to extension.
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The elastic potential energy stored in a spring is given by:
Reasoning: The area under the triangular force-extension graph is $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times F \times}\mathbf{x}$.
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The factor $\mathbf{1/2}$ appears in the elastic potential energy equation because:
Reasoning: Force increases gradually from zero to maximum value, so average force is $\mathbf{F\ /\ 2}$.
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The SI unit of elastic potential energy is:
Reasoning: Elastic potential energy is a form of energy, so its $\mathbf{SI}$ unit is joule.
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A spring is stretched within its proportional limit. The energy stored is:
Reasoning: For elastic deformation, stored energy equals the triangular area under the force-extension graph.
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The equation $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{k}\mathbf{x}^{\mathbf{2}}$ is valid when:
Reasoning: The equation comes from Hooke’s law, $\mathbf{F = k}\mathbf{x}$.
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A spring has a spring constant of $\mathbf{100\ N/m}$ and is stretched by $\mathbf{0.2\ m}$. The elastic energy stored is:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{k}\mathbf{x}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(100)}\mathbf{(0.2)}^{\mathbf{2}}\mathbf{= 2\ J}$
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If the extension of a spring is doubled within the proportional limit, the elastic potential energy becomes:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{k}\mathbf{x}^{\mathbf{2}}$, so energy depends on the square of extension.
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Elastic potential energy is maximum when:
Reasoning: Energy stored increases as extension increases.
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The area under a force-extension graph for an elastic material represents:
Reasoning: The area represents work done, which is stored as elastic potential energy.
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If the force-extension graph is a triangle, the elastic potential energy is:
Reasoning: The area of a triangle is half the product of base and height.
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A spring is stretched from $\mathbf{0\ }$to $\mathbf{5\ cm}$. The force increases uniformly from $\mathbf{0\ }$to $\mathbf{20\ N}$. The energy stored is:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{F}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(20)(0.05) = 0.5\ J}$
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A stiffer spring has a larger spring constant. For the same extension it stores:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}\mathbf{x}}^{\mathbf{2}}$, so larger $\mathbf{k}$ gives greater stored energy.
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The limit of proportionality is important because:
Reasoning: The equation is valid only when force is proportional to extension.
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A force-extension graph has a steeper slope. This indicates:
Reasoning: The slope of the graph represents spring constant.
18 / 40
The energy stored in a stretched spring comes from:
Reasoning: The applied force transfers energy into the spring as elastic potential energy.
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A spring stores $\mathbf{8\ J}$ of energy when stretched by $\mathbf{0.4\ m}$. The force at this extension is:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{F}\mathbf{x}$ $\mathbf{\Longrightarrow}$ $\mathbf{8 =}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(F)(0.4)}$ $\mathbf{\Longrightarrow}$ $\mathbf{F = 40\ N}$
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Which graph feature gives the elastic energy stored?
Reasoning: Energy is determined from the area under the force-extension curve.
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A wire stretched within its elastic limit stores energy because:
Reasoning: Deformation changes the arrangement of particles, storing elastic energy.
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If the spring constant is increased while extension remains constant, elastic energy:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}\mathbf{x}}^{\mathbf{2}}$, so energy increases with $\mathbf{k}$.
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A spring compressed by a distance $\mathbf{x}$ stores energy equal to:
Reasoning: Compression and stretching both store elastic energy according to this equation.
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If a material is stretched beyond its elastic limit, the calculated elastic energy:
Reasoning: Plastic deformation causes some energy loss and permanent change.
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The force-extension graph of an elastic material is useful because it allows calculation of:
Reasoning: The area under the graph gives elastic potential energy.
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The energy stored in a spring depends on:
Reasoning: From $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}\mathbf{x}}^{\mathbf{2}}$, energy depends on $\mathbf{x}^{\mathbf{2}}$.
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A spring stretched to twice its original extension stores:
Reasoning: Energy is proportional to the square of extension.
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The area under the force-extension graph increases if:
Reasoning: Larger force-extension values create a larger area.
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The expression $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{F}\mathbf{x}$ is derived from:
Reasoning: The force-extension graph forms a triangle for Hooke’s law materials.
30 / 40
The elastic potential energy stored in a wire is released when:
Reasoning: Elastic energy is recoverable energy stored during deformation.
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A spring with larger spring constant is harder to stretch because:
Reasoning: Larger $\mathbf{k}$ means greater resistance to deformation.
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A force-extension graph has zero area when:
Reasoning: Without displacement, no work is done and no energy is stored.
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The relationship between elastic potential energy and extension is:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}\mathbf{x}}^{\mathbf{2}}$, so energy varies with square of extension.
34 / 40
A spring of constant $\mathbf{5}\mathbf{0\ N/m}$ is stretched by $\mathbf{0.4\ m}$. Energy stored is:
Reasoning: $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}\mathbf{x}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(50)}\mathbf{(0.4)}^{\mathbf{2}}\mathbf{= 4\ J}$
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Elastic potential energy is not stored when:
Reasoning: No deformation means no elastic energy storage.
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The work done in stretching a spring equals:
Reasoning: Work done is converted into stored elastic energy.
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For a spring obeying Hooke’s law, the force at half the maximum extension is:
Reasoning: Force is directly proportional to extension.
38 / 40
The elastic energy stored in a material depends on:
Reasoning: The force required during deformation determines stored energy.
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If the extension of a spring is reduced to half, the stored energy becomes:
Reasoning: Energy depends on the square of extension.
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The formula $\mathbf{E}_{\mathbf{P}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}\mathbf{x}}^{\mathbf{2}}$ is an application of:
Reasoning: The equation comes from the work done in stretching a Hooke’s law spring.
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