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A steel wire has a length of 2.0 m and a cross-sectional area of 2 x 10⁻⁶ m2. When a force of 200 N is applied, the wire extends by 4 mm. Calculate the tensile stress, tensile strain, and Young's mod

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4 parts
a(i)

1.0×108 Pa\text{1.0} \times 10^{8} \, \text{Pa}

a(ii)

2×10−32 \times 10^{-3}

a(iii)

5.0×1010 Pa5.0 \times 10^{10} \, \text{Pa}

a(v)

58.33 cm58.33 \, \text{cm}

Step-by-step solution

12 steps

Part a(i)· 3 steps

Tensile stress

  1. Identify the given values.

    Force F=200 NF = 200 \, \text{N}
    Cross-sectional area A=2×10−6 m2A = 2 \times 10^{-6} \, \text{m}^2

  2. Use the formula for tensile stress.

    σ=FA\sigma = \frac{F}{A}

  3. Substitute the values and calculate.

    σ=200 N2×10−6 m2\sigma = \frac{200 \, \text{N}}{2 \times 10^{-6} \, \text{m}^2} σ=100×106 N/m2\sigma = 100 \times 10^{6} \, \text{N/m}^2 σ=1.0×108 Pa\sigma = 1.0 \times 10^{8} \, \text{Pa}

    The tensile stress is 1.0×108 Pa\boxed{\text{1.0} \times 10^{8} \, \text{Pa}}.

Part a(ii)

2×10−32 \times 10^{-3}

Tensile strain

  1. Identify the given values and convert units.

    Original length L0=2.0 mL_0 = 2.0 \, \text{m}
    Extension ΔL=4 mm=4×10−3 m\Delta L = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m}

  2. Use the formula for tensile strain.

    ϵ=ΔLL0\epsilon = \frac{\Delta L}{L_0}

  3. Substitute the values and calculate.

    ϵ=4×10−3 m2.0 m\epsilon = \frac{4 \times 10^{-3} \, \text{m}}{2.0 \, \text{m}} ϵ=2×10−3\epsilon = 2 \times 10^{-3}

    The tensile strain is 2×10−3\boxed{2 \times 10^{-3}}.

Part a(iii)

5.0×1010 Pa5.0 \times 10^{10} \, \text{Pa}

Young's modulus

  1. Use the calculated values for tensile stress and tensile strain.

    Tensile stress σ=1.0×108 Pa\sigma = 1.0 \times 10^{8} \, \text{Pa}
    Tensile strain ϵ=2×10−3\epsilon = 2 \times 10^{-3}

  2. Use the formula for Young's modulus.

    Y=σϵY = \frac{\sigma}{\epsilon}

  3. Substitute the values and calculate.

    Y=1.0×108 Pa2×10−3Y = \frac{1.0 \times 10^{8} \, \text{Pa}}{2 \times 10^{-3}} Y=0.5×1011 PaY = 0.5 \times 10^{11} \, \text{Pa} Y=5.0×1010 PaY = 5.0 \times 10^{10} \, \text{Pa}

    Young's modulus is 5.0×1010 Pa\boxed{5.0 \times 10^{10} \, \text{Pa}}.

Part a(iv)

State two applications of elasticity of materials in industry and daily life.

  • Vehicle Tires
  • Suspension systems

Part a(v)

58.33 cm58.33 \, \text{cm}

Using a well labelled diagram, describe a simple experiment to demonstrate Hooke's Law using springs and weights.

To demonstrate Hooke's Law, set up a spring vertically from a rigid support. Attach a ruler alongside the spring to measure its length. Measure the initial length of the spring (L0L_0) when no weight is attached. Then, attach known weights (e.g., 100g, 200g, 300g) one by one to the free end of the spring. For each weight, measure the new length of the spring (LL). Calculate the extension (ΔL=L−L0\Delta L = L - L_0) for each weight. Plot a graph of force (weight) against extension. Hooke's Law states that the extension is directly proportional to the applied force, so the graph should be a straight line passing through the origin, within the elastic limit of the spring.

5) Two springs are connected in series with spring constant of 150 N/m150 \, \text{N/m} and 200 N/m200 \, \text{N/m} respectively. A weight of 50 N50 \, \text{N} is attached at the end of the spring, calculate total extension in cm.

  1. Calculate the equivalent spring constant for springs connected in series.

    Given spring constants: k1=150 N/mk_1 = 150 \, \text{N/m}, k2=200 N/mk_2 = 200 \, \text{N/m}.
    For springs in series, the reciprocal of the equivalent spring constant is the sum of the reciprocals of individual spring constants:

    1keq=1k1+1k2\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} 1keq=1150 N/m+1200 N/m\frac{1}{k_{eq}} = \frac{1}{150 \, \text{N/m}} + \frac{1}{200 \, \text{N/m}} 1keq=4600 N/m+3600 N/m\frac{1}{k_{eq}} = \frac{4}{600 \, \text{N/m}} + \frac{3}{600 \, \text{N/m}} 1keq=7600 N/m\frac{1}{k_{eq}} = \frac{7}{600 \, \text{N/m}} keq=6007 N/mk_{eq} = \frac{600}{7} \, \text{N/m} keq≈85.714 N/mk_{eq} \approx 85.714 \, \text{N/m}

  2. Calculate the total extension using Hooke's Law.

    Force F=50 NF = 50 \, \text{N}

    F=keq×xF = k_{eq} \times x x=Fkeqx = \frac{F}{k_{eq}} x=50 N6007 N/mx = \frac{50 \, \text{N}}{\frac{600}{7} \, \text{N/m}} x=50×7600 mx = \frac{50 \times 7}{600} \, \text{m} x=350600 mx = \frac{350}{600} \, \text{m} x=712 mx = \frac{7}{12} \, \text{m} x≈0.5833 mx \approx 0.5833 \, \text{m}

  3. Convert the total extension to centimeters.

    x=0.5833 m×100 cm/mx = 0.5833 \, \text{m} \times 100 \, \text{cm/m} x=58.33 cmx = 58.33 \, \text{cm}

    The total extension is 58.33 cm\boxed{58.33 \, \text{cm}}.

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