Question

When a block a mass M is suspended by a long wire of length L, the elastic potential potential energy stored in the wire is 12 × stress × strain × volume. Show that it is equal to 12 Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining 12 Mgl energy go?

Answer 1

Explanation:

• Stress =$\frac{M g}{A}$ (Where A is considered as the wire's cross-sectional area)

• Strain $=\frac{l}{L}$  

• Since the energy of elastic potential

$U=\frac{1}{2}(\text {Stress}) \times(\text {Strain}) \times \text { (Volume) } U=1 / 2 *\left(\frac{M g}{A}\right) \times\left(\frac{l}{L}\right) \times(A L)=\frac{1}{2} M g l$

• Only when the load is suddenly released from the initial position i.e. when the elongation is zero will the loss in gravitational potential energy be $\mathrm{Mgl}$

The elastic energy which is contained in the wire =$=\frac{1}{2} \mathrm{Mgl}$

• If the elongation is l the remainder of the gravitational energy potential $\mathrm{Mgl}^{\frac{1}{2}} \mathrm{Mg} |=\frac{1}{2} \mathrm{Mgl}$  is converted to the kinetic energy. The mass M will have a certain speed, say v, at the time elongation is l.

So $\frac{1}{2} M v^{2}=\frac{1}{2} M g l_{v^{2}}=g l v=\sqrt{(g l)}$  

• This point as a mean position the mass will be vertically in a simple harmonic motion. Slowly, by dissipating that energy in the form of heat, it will come to rest at the mean position.