5.11: End-of-Chapter Problem Set

End-of-Chapter Problem Set

The problems below combine unit conversion, SI prefixes, dimensional analysis, computational-unit discipline, and order-of-magnitude reasoning. Show your unit cancellations whenever a conversion is required. A numerical answer without units is incomplete.

Reference — Units and Symbols Used in This Problem Set

You have seen these units earlier in the chapter. Use this list if you need a reminder:

• ft = foot, a unit of length
• mi = mile, a unit of length
• mph = miles per hour, a unit of speed
• m/s = meters per second, a unit of speed
• lb_f = pound-force, a unit of force
• N = newton, the SI unit of force
• L = liter, a unit of volume
• Pa = pascal, a unit of pressure or stress
• A = ampere, a unit of electric current
• V = volt, a unit of voltage
• Ω = ohm, a unit of electrical resistance
• W = watt, a unit of power

Remember that prefixes such as μ, m, k, M, and G change the scale of the unit.

Unit Conversions

1. Convert 150 ft (feet) to meters.
2. Convert 120 lb_f (pound-force) to newtons.
3. Convert 35 mph (miles per hour) to m/s (meters per second). Use \(1\text{ mi}=1609.34\text{ m}\) and \(1\text{ h}=3600\text{ s}\).
4. Convert 0.0025 m³ to liters. Use \(1 \text{ L} = 10^{-3} \text{ m}^3\).
5. Convert 85 kPa (kilopascals) to Pa (pascals).
6. Convert 3.2 GPa (gigapascals) to MPa (megapascals).
7. Convert 450 μA (microamperes) to mA (milliamperes).
8. Convert 330 km/h (kilometers per hour) to ft/s (feet per second). Use \(1\text{ m}=3.28084\text{ ft}\).

Dimensional Verification

For each problem, compare the dimensions on both sides of the equation. Remember that dimensional consistency tells you whether an equation could be correct; it does not prove that it is the correct physical model.

9. Verify the dimensional consistency of \(F = ma\), where \(F\) is force, \(m\) is mass, and \(a\) is acceleration.
10. Verify the dimensional consistency of \(P = I^2R\), where \(P\) is electrical power, \(I\) is current, and \(R\) is resistance. You may use \(R=V/I\) and \(1\text{ W}=1\text{ V}\cdot1\text{ A}\).
11. Determine whether \(v = at^2\) is dimensionally valid, where \(v\) is velocity, \(a\) is acceleration, and \(t\) is time. If it is not valid, identify the dimensional mismatch.
12. Suppose \(E\) represents kinetic energy. Determine whether \(E = mv\) is dimensionally valid. If not, what familiar kinetic-energy equation has the correct dimensions?

Applied Problems

13. A 12 V source is connected to a 330 Ω resistor. Calculate the current in mA and the electrical power in mW. Use \(I=V/R\) and \(P=VI\).
14. A student reports a cable tension of 850 kg. Explain what is wrong with this result using the error terminology from Chapter 4. What type of physical quantity is tension, and what SI unit should be used?
15. A 24 V source is connected to a 4.7 kΩ resistor in a spreadsheet model. The model expects resistance in ohms, but the student enters R = 4.7. Calculate the current the model will compute, compare it with the correct current, and determine the factor of error. Use \(I=V/R\).
16. A pressure sensor reports \(2.3 \times 10^{-2}\) GPa. Convert this value to kPa.
17. [Challenge] A container holds 0.75 m³ of water. The density of water is \(1000\text{ kg/m}^3\). Calculate:
    1. the mass of the water in kg,
    2. the weight of the water in N, and
    3. the weight in lb_f.
    Use \(W=mg\), \(g=9.81\text{ m/s}^2\), and \(1\text{ lb}_{\text f}=4.448\text{ N}\). Show all unit cancellations explicitly.

Computational and Scale Checks

18. A spreadsheet column is labeled only Pressure. The values entered are 2.4, 3.1, and 4.8. Explain why these entries are incomplete engineering information. Rewrite the column heading so that the model clearly expects pressure in MPa.
19. A sensor circuit is expected to produce a current on the order of milliamperes. A calculation reports \(3.6\text{ A}\). The units are correct. Explain why the result should still be investigated, and identify one likely type of error based on the material in this chapter.

Answer Key - click to expand

Unit Conversions

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

Dimensional Verification

9. \(F=ma\) is dimensionally consistent. \[ [ma] = M\left(\frac{L}{T^2}\right) = \frac{ML}{T^2} \] Force also has dimensions \(\dfrac{ML}{T^2}\), so both sides match.
10. \(P=I^2R\) is dimensionally consistent. \[ I^2R = I^2\left(\frac{V}{I}\right) = IV \] Since ampere × volt is watt, the right side has units of power.
11. \(v=at^2\) is dimensionally invalid. \[ [at^2] = \frac{L}{T^2}T^2 = L \] The right side has dimensions of length, while velocity has dimensions \(L/T\). A dimensionally valid constant-acceleration relationship is \(v=at\) when the initial velocity is zero.
12. \(E=mv\) is not dimensionally valid for kinetic energy. \[ [mv] = M\frac{L}{T} \] These are the dimensions of momentum. Kinetic energy is: \[ E_k=\frac{1}{2}mv^2 \] which has dimensions: \[ M\frac{L^2}{T^2} \]

Applied Problems

13. Current: \[ I = \frac{V}{R} = \frac{12\text{ V}}{330\Omega} = 0.03636\text{ A} = 36.4\text{ mA} \]

    Power:

    \[ P = VI = (12\text{ V})(0.03636\text{ A}) = 0.436\text{ W} = 436\text{ mW} \]
14. This is a unit or dimensional error. Tension is a force, while kilograms are units of mass. The appropriate SI unit for tension is the newton (N).

    The statement “850 kg” by itself does not provide enough information to determine the correct numerical tension. It tells us that the reported unit is wrong.

    If 850 kg were instead the mass of an object whose weight was producing the tension, then additional modeling could give:

    \[ W = mg = (850\text{ kg}) \left( 9.81\frac{\text{m}}{\text{s}^2} \right) \approx 8.34\text{ kN} \]

    But that conclusion requires the additional assumption that the 850 kg value represents mass.

15. The incorrect model interprets the resistance as \(4.7\Omega\): \[ I_{\text{model}} = \frac{24\text{ V}}{4.7\Omega} = 5.11\text{ A} \]

    The correct resistance is:

    \[ 4.7\text{ k}\Omega = 4700\Omega \]

    Therefore:

    \[ I_{\text{correct}} = \frac{24\text{ V}}{4700\Omega} = 0.00511\text{ A} = 5.11\text{ mA} \]

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

16. Since: \[ 1\text{ GPa}=10^6\text{ kPa} \] then: \[ 2.3\times10^{-2}\text{ GPa} \times \frac{10^6\text{ kPa}}{1\text{ GPa}} = 2.3\times10^4\text{ kPa} = 23{,}000\text{ kPa} \]
17. Mass: \[ m = 0.75\text{ m}^3 \times \frac{1000\text{ kg}}{1\text{ m}^3} = 750\text{ kg} \]

    Weight in newtons:

    \[ W = (750\text{ kg}) \left( 9.81\frac{\text{m}}{\text{s}^2} \right) = 7357.5\text{ N} \]

    Weight in pounds-force:

    \[ 7357.5\text{ N} \times \frac{1\text{ lb}_{\text f}}{4.448\text{ N}} = 1654\text{ lb}_{\text f} \]

Computational and Scale Checks

18. The values 2.4, 3.1, and 4.8 are incomplete because the spreadsheet does not identify the unit associated with the pressure values.

    A clearer heading would be:

    Pressure, P (MPa)

    This tells anyone using the model that the numerical entries are expected to be in megapascals.

19. A current expected to be on the order of milliamperes should typically be near \(10^{-3}\text{ A}\). A result of \(3.6\text{ A}\) is on the order of \(10^0\text{ A}\).

    The result is therefore about three orders of magnitude larger than expected. The units may be correct, but the scale is suspicious.

    One likely cause is an SI-prefix error, such as entering a resistance in kΩ as though it were in Ω. The result should be checked before being accepted.