Question

A wire suspended vertically from one end is stretched by attaching a weight of 20 N to the lower end. The weight stretches the wire by 1mm .How much energy is gained by the wire

Answer 1

Answer on Question #41139, Physics, Mechanics | Kinematics | Dynamics

Question:

A wire suspended vertically from one end is stretched by attaching a weight of 20 N to the lower end. The weight stretches the wire by 1mm. How much energy is gained by the wire

Answer:

Energy of wire deformation equals:

$$E = \frac {k x ^ {2}}{2}$$

where $k$ is force constant, $x$ is deformation.

Force constant equals:

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

where $F$ is force.

Therefore:

$$E = \frac {F}{2 x} x ^ {2} = \frac {F x}{2} = \frac {20 N 0.001 m}{2} = 0.01 J$$

Answer: 0.01 J

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