An engineer is designing a metal part for a machine - Edexcel - A-Level Physics - Question 14 - 2023 - Paper 1

To find the Young's modulus (E), we can use the formula:

$E = \frac{\text{Stress}}{\text{Strain}}$

From the stress-strain graph, we can determine the slope (stress/strain) in the linear region.

Using a stress of about $4 \times 10^{6}$ Pa (from the graph corresponding to a strain of $2 \times 10^{-3}$), we have:

• Stress = $4 \times 10^{6}$ Pa
• Strain = $2 \times 10^{-3}$ (this is the change in length divided by the original length)

Thus,

$E = \frac{4 \times 10^{6}}{2 \times 10^{-3}} = 2 \times 10^{11} \text{ Pa}$

First, we need to convert the maximum allowable compression of 0.60 mm to meters:

$0.60 \text{ mm} = 0.60 \times 10^{-3} ext{ m}$

Using the formula for deformation (compression) under a load:

$\Delta L = \frac{F L}{A E}$

Where:

• $F = 9.5 \times 10^{5}$ N (force)
• $L = 0.84$ m (original length)
• $A = 4.8 \times 10^{-4}$ m² (cross-sectional area)
• $E = 2 \times 10^{11}$ Pa (Young's modulus)

Substituting the values:

$\Delta L = \frac{(9.5 \times 10^{5}) (0.84)}{(4.8 \times 10^{-4}) (2 \times 10^{11})}$

Calculating this gives:

$\Delta L \approx 0.00004 \text{ m} = 0.04 ext{ mm}$

Since 0.04 mm < 0.60 mm, the metal is suitable for the part.