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12a (i). What is meant by resultant force? (ii). Consider four forces acting at a point as shown in figure 3. Find the magnitude and direction of the resultant force.

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2 parts
a

0.015 J\text{0.015 J}

b

0.0005 J\text{0.0005 J}

Step-by-step solution

4 steps

Here's the solution to question 13b:

Part a· 2 steps

  1. Identify the given values for part (i).

    The force applied to the wire is F=100 NF = 100 \, \text{N}.
    The extension produced is e=0.3 mm=0.3×10−3 me = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m}.

  2. Calculate the potential energy stored in the wire.

    The potential energy (elastic potential energy) stored in a stretched wire is given by the formula:

    PE=12FePE = \frac{1}{2} F e

    Substitute the given values:

    PE=12×100 N×(0.3×10−3 m)PE = \frac{1}{2} \times 100 \, \text{N} \times (0.3 \times 10^{-3} \, \text{m}) PE=50×0.3×10−3 JPE = 50 \times 0.3 \times 10^{-3} \, \text{J} PE=15×10−3 JPE = 15 \times 10^{-3} \, \text{J} PE=0.015 JPE = 0.015 \, \text{J}

    The potential energy in the wire is 0.015 J\boxed{\text{0.015 J}}.

Part b

0.0005 J\text{0.0005 J}

  1. Identify the given values for part (ii).

    The length of the wire is L′=2 mL' = 2 \, \text{m}.
    The cross-sectional area is A′=1 mm2=1×(10−3)2 m2=1×10−6 m2A' = 1 \, \text{mm}^2 = 1 \times (10^{-3})^2 \, \text{m}^2 = 1 \times 10^{-6} \, \text{m}^2.
    The extension is e′=0.1 mm=0.1×10−3 me' = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}.
    The Young's Modulus of the wire is Y=2.0×1011 Nm−2Y = 2.0 \times 10^{11} \, \text{Nm}^{-2}.

  2. Calculate the work done in stretching the wire.

    The work done in stretching a wire is equal to the elastic potential energy stored, which can be expressed using Young's Modulus:

    W=12A′YL′(e′)2W = \frac{1}{2} \frac{A'Y}{L'} (e')^2

    Substitute the given values:

    W=12×(1×10−6 m2)×(2.0×1011 Nm−2)2 m×(0.1×10−3 m)2W = \frac{1}{2} \times \frac{(1 \times 10^{-6} \, \text{m}^2) \times (2.0 \times 10^{11} \, \text{Nm}^{-2})}{2 \, \text{m}} \times (0.1 \times 10^{-3} \, \text{m})^2 W=12×2.0×105 N2×(1×10−2×10−6 m2)W = \frac{1}{2} \times \frac{2.0 \times 10^{5} \, \text{N}}{2} \times (1 \times 10^{-2} \times 10^{-6} \, \text{m}^2) W=12×(1.0×105 N)×(1×10−8 m2)W = \frac{1}{2} \times (1.0 \times 10^{5} \, \text{N}) \times (1 \times 10^{-8} \, \text{m}^2) W=0.5×10−3 JW = 0.5 \times 10^{-3} \, \text{J} W=0.0005 JW = 0.0005 \, \text{J}

    The work done in stretching the wire is 0.0005 J\boxed{\text{0.0005 J}}.

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