11 Example 2.16

Kinetic Energy Dimensional Analysis

This simulation demonstrates how to use dimensional analysis to evaluate kinetic energy formulas, based on Example 2.16 from your Physics textbook.

Correct dimensions of kinetic energy (K):

[K] = [M L² T⁻²] (same as joules)

[M]

Mass

[L]

Length

[T]

Time

Evaluate These Kinetic Energy Formulas:

(a) K = m²v³

Dimensions of m²: [M]² = [M²]

Dimensions of v³: [L T⁻¹]³ = [L³ T⁻³]

Dimensions of m²v³: [M²] × [L³ T⁻³] = [M² L³ T⁻³]

Does NOT match [M L² T⁻²]

(b) K = (1/2)mv²

Dimensions of m: [M]

Dimensions of v²: [L T⁻¹]² = [L² T⁻²]

Dimensions of mv²: [M] × [L² T⁻²] = [M L² T⁻²]

The constant (1/2) is dimensionless

Matches [M L² T⁻²]

Dimensions of m: [M]

Dimensions of a: [L T⁻²]

Dimensions of ma: [M] × [L T⁻²] = [M L T⁻²]

Does NOT match [M L² T⁻²]

(d) K = (3/16)mv²

Dimensions of m: [M]

Dimensions of v²: [L T⁻¹]² = [L² T⁻²]

Dimensions of mv²: [M] × [L² T⁻²] = [M L² T⁻²]

The constant (3/16) is dimensionless

Matches [M L² T⁻²]

(e) K = (1/2)mv² + ma

First term: (1/2)mv² → [M L² T⁻²]

Second term: ma → [M L T⁻²]

Cannot add quantities with different dimensions

Conclusion:

Formulas (b) and (d) have the correct dimensions for kinetic energy.

Formulas (a), (c), and (e) can be ruled out based on dimensional analysis.

Note: While (b) and (d) are both dimensionally correct, (b) is the physically correct formula for kinetic energy.

Example 2.16 Solution:

We can rule out options (a), (c), and (e) because:

(a) has dimensions [M² L³ T⁻³] which don't match energy
(c) has dimensions [M L T⁻²] (same as force, not energy)
(e) tries to add two terms with different dimensions

Options (b) and (d) both have the correct dimensions [M L² T⁻²] for energy.

11 Example 2.16

Kinetic Energy Dimensional Analysis

Evaluate These Kinetic Energy Formulas:

Conclusion:

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