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 #41143, Physics, Mechanics | Kinematics | Dynamics

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

Solution:

Elastic potential energy is an energy stored as a result of deformation of an elastic object, such as the stretching of a wire. It is equal to the work done to stretch the spring, which depends upon the spring constant $k$ as well as the distance stretched. According to Hooke's law, the force required to stretch the wire will be directly proportional to the amount of stretch.

The force has the form

$$F = -kx$$

The work done to stretch the spring a distance $x$ is

$$W = PE = \frac{kx^2}{2}.$$

Thus, the energy stored in the stretch wire is

$$PE = \frac{1}{2}Fx$$

where

$PE$ = elastic strain energy in joules (J)

$F$ = force in newtons (N)

$x$ = change in length in meters (m)

So,

$$PE = \frac{1}{2} \cdot 20 \cdot 0.001 = 0.01\,\mathrm{J}$$

Answer. $PE = 0.01\,\mathrm{J}$

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