Question

steel wire of length 2.5 m and area of 5
cross-section 2-5 x 10 -6M2 is suspended
from a torsion head. A 5 kg weight is
suspended at its free-end.
Calculate the work done on the wire.
Take Y= 2 x 1011 N in 2 .

Answer 1

Steel wire of length $2.5\,\mathrm{m}$ and area of cross-section $2.5 \times 10^{-6}\,\mathrm{m}^2$ is suspended from a torsion head. A $5\,\mathrm{kg}$ weight is suspended at its free-end. Calculate the work done on the wire.

Take $= 2 \times 10^{11} \frac{\mathrm{N}}{\mathrm{m}^2}$.

Solution

Work done in the wire of length $l$ to a length $l + \Delta l$ by force $F$ is

$$W = \frac{1}{2} \sigma \varepsilon V,$$

where $\sigma = \frac{F}{A}$ — stress, $A$ — area, $\varepsilon = \frac{\Delta l}{l}$ — strain, $V$ — volume.

$$Y = \frac{\sigma}{\varepsilon} \rightarrow \varepsilon = \frac{\sigma}{Y}.$$

where $Y$ — Young's modulus.

So

$$W = \frac{1}{2} \sigma \cdot \frac{\sigma}{Y} \cdot V = \frac{\sigma^2 V}{2 Y}.$$

A force $F$ in that case is weight of $5\,\mathrm{kg}$:

$$F = m g,$$

where $g$ — acceleration due to the gravity.

Now we have

$$W = \frac{\left(\frac{m g}{A}\right)^2 A \cdot l}{2 Y} = \frac{m^2 \cdot g^2 \cdot l}{2 \cdot A \cdot Y}.$$ $$W = \frac{5^2 \,\mathrm{kg}^2 \cdot 9.8^2 \left(\frac{\mathrm{N}}{\mathrm{kg}}\right)^2 \cdot 2.5\,\mathrm{m}}{2 \cdot 2.5 \cdot 10^{-6}\,\mathrm{m}^2 \cdot 2 \cdot 10^{11} \frac{\mathrm{N}}{\mathrm{m}^2}} = 6 \cdot 10^{-3} J.$$

Answer: $6 \cdot 10^{-3} J$.