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Engineering LibreTexts

5.11: End-of-Chapter Problem Set

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    End-of-Chapter Problem Set

    Unit Conversions

    1. Convert 150 ft to meters.
    2. Convert 120 lb\(_\text{f}\) to newtons.
    3. Convert 35 mph to m/s.
    4. Convert 0.0025 m³ to liters. (Note: \(1 \text{ L} = 10^{-3} \text{ m}^3\))
    5. Convert 85 kPa to Pa.
    6. Convert 3.2 GPa to MPa.
    7. Convert 450 μA to mA.
    8. Convert 330 km/h to ft/s. (Multi-step: use both metric and imperial conversions.)

    Dimensional Verification

    1. Verify the dimensional consistency of \(F = ma\).
    2. Verify the dimensional consistency of \(P = I^2 R\).
    3. Determine whether \(v = at^2\) is dimensionally valid. If not, what is wrong?
    4. Determine whether \(E = mv\) is dimensionally valid. If not, what is the correct form?

    Applied Problems

    1. A 12 V source is connected to a 330 Ω resistor. Calculate the current in mA and the power in mW.
    2. A student calculates a cable tension of 850 kg. Identify the error type from Chapter 4's error taxonomy and state the correct unit.
    3. A 24 V source is connected to a 4.7 kΩ resistor in a spreadsheet model. The student enters \(R = 4.7\). Calculate the current the model will compute, compare it to the correct current, and state the factor of error.
    4. A pressure sensor reports \(2.3 \times 10^{-2}\) GPa. Convert this to kPa.
    5. [Challenge] A container holds 0.75 m³ of water. Density of water = 1000 kg/m³. Calculate the mass in kg and weight in N and lb\(_\text{f}\). Show all unit cancellations explicitly.
    Answer Key - click to expand

    Unit Conversions

    1. \(150 \text{ ft} \times \dfrac{0.3048 \text{ m}}{1 \text{ ft}} = 45.72 \text{ m}\)
    2. \(120 \text{ lb}_\text{f} \times \dfrac{4.448 \text{ N}}{1 \text{ lb}_\text{f}} = 533.8 \text{ N}\)
    3. \(35 \text{ mph} \times \dfrac{1609.34 \text{ m}}{1 \text{ mi}} \times \dfrac{1 \text{ hr}}{3600 \text{ s}} = 15.65 \text{ m/s}\)
    4. \(0.0025 \text{ m}^3 \times \dfrac{1 \text{ L}}{10^{-3} \text{ m}^3} = 2.5 \text{ L}\)
    5. \(85 \text{ kPa} \times \dfrac{1000 \text{ Pa}}{1 \text{ kPa}} = 85{,}000 \text{ Pa}\)
    6. \(3.2 \text{ GPa} \times \dfrac{1000 \text{ MPa}}{1 \text{ GPa}} = 3200 \text{ MPa}\)
    7. \(450 \ \mu\text{A} \times \dfrac{1 \text{ mA}}{1000 \ \mu\text{A}} = 0.45 \text{ mA}\)
    8. \(330 \dfrac{\text{km}}{\text{h}} \times \dfrac{1000 \text{ m}}{1 \text{ km}} \times \dfrac{3.28084 \text{ ft}}{1 \text{ m}} \times \dfrac{1 \text{ h}}{3600 \text{ s}} = 300.7 \dfrac{\text{ft}}{\text{s}}\)

    Dimensional Verification

    1. \(F = ma\) is dimensionally consistent. Since \([m][a] = M \cdot \dfrac{L}{T^2} = \dfrac{ML}{T^2}\), and force has dimensions \(\dfrac{ML}{T^2}\), the equation is valid.
    2. \(P = I^2R\) is dimensionally consistent. Since \(R = \dfrac{V}{I}\), then \(I^2R = I^2\left(\dfrac{V}{I}\right) = IV\). Current times voltage gives power, so the result is watts.
    3. \(v = at^2\) is not dimensionally valid. The right side has dimensions \([a][t^2] = \dfrac{L}{T^2} \cdot T^2 = L\), which is length, not velocity. A valid constant-acceleration form is \(v = at\), assuming initial velocity is zero.
    4. \(E = mv\) is not dimensionally valid for energy. The right side has dimensions \([m][v] = M \cdot \dfrac{L}{T} = \dfrac{ML}{T}\), which is momentum, not energy. The kinetic energy form is \(E = \dfrac{1}{2}mv^2\).

    Applied Problems

    1. Current: \(I = \dfrac{V}{R} = \dfrac{12 \text{ V}}{330 \ \Omega} = 0.03636 \text{ A} = 36.4 \text{ mA}\).

      Power: \(P = VI = (12)(0.03636) = 0.436 \text{ W} = 436 \text{ mW}\).

    2. This is a unit or dimensional error. Tension is a force, so the correct SI unit is newtons, \(\text{N}\), not kilograms.

      If \(850 \text{ kg}\) represents an equivalent mass, the corresponding weight force would be \(W = mg = (850 \text{ kg})\left(9.81 \dfrac{\text{m}}{\text{s}^2}\right) = 8338.5 \text{ N} \approx 8.34 \text{ kN}\).

    3. The spreadsheet model computes \(I_\text{model} = \dfrac{24 \text{ V}}{4.7 \ \Omega} = 5.11 \text{ A}\).

      The correct resistance is \(4.7 \text{ k}\Omega = 4700 \ \Omega\), so the correct current is \(I_\text{correct} = \dfrac{24 \text{ V}}{4700 \ \Omega} = 0.00511 \text{ A} = 5.11 \text{ mA}\).

      The model current is \(1000\) times too large.

    4. \(2.3 \times 10^{-2} \text{ GPa} \times \dfrac{10^6 \text{ kPa}}{1 \text{ GPa}} = 2.3 \times 10^4 \text{ kPa} = 23{,}000 \text{ kPa}\).
    5. Mass: \(m = 0.75 \text{ m}^3 \times \dfrac{1000 \text{ kg}}{1 \text{ m}^3} = 750 \text{ kg}\).

      Weight in newtons: \(W = 750 \text{ kg} \times 9.81 \dfrac{\text{m}}{\text{s}^2} = 7357.5 \text{ N}\).

      Weight in pounds-force: \(7357.5 \text{ N} \times \dfrac{1 \text{ lb}_\text{f}}{4.448 \text{ N}} = 1654 \text{ lb}_\text{f}\).

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