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 four. The wire was elongated by \(1 \mathrm{~mm}\). What is the stored in the wire due to stretching ? (A) Zero (B) \(0.05\) Joule (C) 100 Joule (D) 500 Joule

Short Answer

Expert verified
(B) \(0.05\) Joule

Step by step solution

01

Calculate the force exerted by the weight

With the given mass, we can find the force this mass exerts on the wire due to gravity. The force can be calculated using the formula: \(F = mg\) where m is the mass (10 kg) and g is the acceleration due to gravity (approximated as \(9.81 \mathrm{~m/s^2}\)). \(F = (10\mathrm{~kg})(9.81 \mathrm{~m/s^2})\) \(F = 98.1 \mathrm{~N}\) (Newtons) Step 2: Calculate the spring constant (k)
02

Calculate the spring constant (k)

Using Hooke's law, we have the force and elongation relation as: \(F = kx\) We can rewrite this equation for the spring constant (k) as: \(k = \frac{F}{x}\) where x is the elongation (1 mm, or 0.001 m). We plug in the force and elongation values to find k. \(k = \frac{98.1 \mathrm{~N}}{0.001 \mathrm{~m}}\) \(k = 98100 \mathrm{~N/m}\) Step 3: Calculate the potential energy stored in the wire
03

Calculate the potential energy stored in the wire

We can now use the formula for the potential energy of a stretched wire: \(U = \frac{1}{2}kx^2\) and plug in the spring constant (k) and elongation (x) from previous steps: \(U = \frac{1}{2}(98100 \mathrm{~N/m})(0.001 \mathrm{~m})^2\) \(U = \frac{1}{2}(98100 \mathrm{~N/m})(0.000001 \mathrm{~m^2})\) \(U = 0.04905 \mathrm{~J}\) The stored potential energy in the wire is approximately 0.04905 Joules. We can round it to 0.05 Joules. Therefore, the correct answer is: (B) \(0.05\) Joule

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