Question

a 5 meter long wire is fixed to the ceiling. a weight of 10kg is hung at the lower end and is 1 meter above the floor . the wire was elongated by 1mm . the energy stored in the wire due to stretching is

Answer 1

Answer on Question #62918, Physics / Mechanics | Relativity

A 5-meter long wire is fixed to the ceiling. A weight of $10\mathrm{kg}$ is hung at the lower end and is 1 meter above the floor. The wire was elongated by $1\mathrm{mm}$. The energy stored in the wire due to stretching is.

Solution:

The energy of the deformed wire is determined by the formula

$$w = \frac{1}{2} k l^{2}$$

Hooke's law

$$F = k l$$

Whence

$$k = \frac{F}{l}$$

Then

$$w = \frac{1}{2} \times \frac{F}{l} \times l^{2}$$

Where

$$F = m g$$

Finally

$$w = \frac{1}{2} \times m g \times l$$ $$w = \frac{1}{2} \times 10 k g \times 9.8 \, \mathrm{m/s^2} \times 0.001 \, \mathrm{m} = 0.05 \, \mathrm{joule}$$

Answer: 0.05 joule

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